Welcome to our blog on calculations, questions, and answers related to Temperature and Thermal Equilibrium, Thermometers and Temperature Scales, Thermal Expansion, Heat Transfer, Specific Heat, Calorimetry, Phase Changes, and Mechanisms of Heat Transfer.
Temperature is an essential concept in understanding the behavior of matter. It is the measure of the average kinetic energy of the particles in a system. Thermometers are devices used to measure temperature, and they use different temperature scales, such as Celsius, Fahrenheit, and Kelvin.
Thermal expansion is the tendency of matter to change its size, shape, and volume in response to changes in temperature. This phenomenon is applied in many practical applications, such as the construction of bridges and buildings.
Heat transfer is the process of moving energy from one place to another as a result of a difference in temperature. Specific heat is the amount of heat energy required to raise the temperature of a substance by one degree Celsius or Kelvin. Calorimetry is the measurement of heat energy transfer during a chemical reaction.
Phase changes occur when a substance changes from one state of matter to another. The most common phases are solid, liquid, and gas. The mechanisms of heat transfer include conduction, convection, and radiation.
In this blog, we will explore various calculations, questions, and answers related to these topics. Our aim is to help readers develop a deeper understanding of these concepts and their applications in diverse fields. Whether you are a student, researcher, or simply curious about science, you will find valuable insights and information in our blog. Stay tuned for our upcoming posts!
No1
A person taking a reading of the temperature in a freezer in Celsius makes two mistakes: first omitting the negative sign and then thinking the temperature is Fahrenheit. That is, the person reads -X °C as x°F.. Oddly enough, the result is the correct Fahrenheit temperature. What is the original Celsius reading? Round your answer to three significant figures.
SOLUTION
Since the person reads -X °C as x°F and the result is the correct Fahrenheit temperature, we can assume that the original Celsius reading is -X degrees.
To convert -X degrees Celsius to Fahrenheit, we use the formula:
F = (C × 9/5) + 32
Substituting -X for C, we have:
x = (-X × 9/5) + 32
Rearranging the equation:
9X/5 = 32 - x
9X = 5(32 - x)
9X = 160 - 5x
9X + 5x = 160
14X = 160
X = 160/14
X = 11.428571...
Rounding to three significant figures, the original Celsius reading is approximately -11.4 degrees Celsius.
Answer -11.4°C
No2
Global warming will produce rising sea levels partly due to melting ice caps and partly doe to the expansion of water as average ocean temperatures rise. To get some idea of the size of this effect, calculate the change in length of a column of water 1.00 km high for a temperature increase of 100 °C. Assume the column is not free to expand sideways. As a model of the ocean, that is a reasonable approximation, as only parts of the ocean very close to the surface can expand sideways onto land, and only to a limited degree. As another approximation, neglect the fact that ocean warming is not uniform with depth.
SOLUTION
To calculate the change in length of a column of water, we can use the coefficient of thermal expansion, which measures how much a material expands or contracts with a change in temperature.
The coefficient of thermal expansion for water is approximately 0.00021 per °C. This means that for every degree Celsius increase in temperature, water expands by 0.00021 times its original length.
In this case, we have a 1.00 km high column of water and a temperature increase of 100 °C. To find the change in length, we multiply the original length of the column (1.00 km) by the coefficient of thermal expansion (0.00021) and the temperature increase (100):
Change in length = Original length * Coefficient of thermal expansion * Temperature increase
Change in length = (1.00 km) * (0.00021/°C) * (100 °C)
Converting km to m:
1.00 km = 1000 m
Change in length = (1000 m) * (0.00021/°C) * (100 °C)
Change in length = 1000 * 0.00021 * 100 m
Change in length = 21 m
Therefore, for a temperature increase of 100 °C, the column of water would expand by approximately 21 meters.
No3
Suppose a meter stick made of steel and one made of aluminum are the same length at 0°C. What is their difference in length at 220°C?
SOLUTION
To calculate the change in length of a column of water, we can use the coefficient of thermal expansion, which measures how much a material expands or contracts with a change in temperature.
The coefficient of thermal expansion for water is approximately 0.00021 per °C. This means that for every degree Celsius increase in temperature, water expands by 0.00021 times its original length.
In this case, we have a 1.00 km high column of water and a temperature increase of 100 °C. To find the change in length, we multiply the original length of the column (1.00 km) by the coefficient of thermal expansion (0.00021) and the temperature increase (100):
Change in length = Original length * Coefficient of thermal expansion * Temperature increase
Change in length = (1.00 km) * (0.00021/°C) * (100 °C)
Converting km to m:
1.00 km = 1000 m
Change in length = (1000 m) * (0.00021/°C) * (100 °C)
Change in length = 1000 * 0.00021 * 100 m
Change in length = 21 m
Therefore, for a temperature increase of 100 °C, the column of water would expand by approximately 21 meters.
No4
Most cars have a coolant reservoir to catch radiator fluid that may overflow when the engine is hot. A radiator is made of copper and is filled to it's 16.0-L capacity when at 100 °C. What volume of radiator fluid will overflow
SOLUTION
To calculate the volume of radiator fluid that will overflow, we need to consider the thermal expansion of copper and the temperature difference between the filled radiator and its capacity temperature.
No5
The height of the Washington Monument is measured be 170.00 m on a day when the temperature is 350°C. What will its height be an a day when the temperature falls to -10.0°C? Although the momamem is made ut limestone, assume that its coefficient of thermal expansion is the same as that of marble. Give your answer
SOLUTION
The coefficient of thermal expansion of marble is 5.5 x 10^-6 /°C.
First, we need to calculate the change in height of the Washington Monument due to the change in temperature.
ΔH = αLΔT
where ΔH is the change in height, α is the coefficient of thermal expansion, L is the original height, and ΔT is the change in temperature.
ΔH = (5.5 x 10^-6 /°C) x 170.00 m x (350°C - (-10.0°C)) ΔH = 0.10 m
So, the height of the Washington Monument will decrease by 0.10 m when the temperature drops from 350°C to -10.0°C.
Therefore, the new height of the Washington Monument will be:
New height = 170.00 m - 0.10 m = 169.90 m
Rounded to the appropriate significant figures, the new height of the Washington Monument is 169.9 m.
No6
What is the change in length at a 3.00-cm-long column of mercury if its temperature changes from 37.0°C to 400°C, assuming the mercury is constrained to a cylinder but unconstrained in length? Your answer will show why thermometers contain bulbs at the bottom instead of simple columns of liquid.
SOLUTION
The coefficient of thermal expansion of mercury is 1.82 x 10^-4 /°C.
We can use the formula for linear thermal expansion:
ΔL = αLΔT
where ΔL is the change in length, α is the coefficient of thermal expansion, L is the original length and ΔT is the change in temperature.
For a column of mercury that is unconstrained in length, the change in length will be equal to the change in height of the column.
ΔL = αLΔT
ΔL = (1.82 x 10^-4 /°C) x (3.00 cm) x (400°C - 37.0°C)
ΔL = 1.74 cm
Therefore, the change in length of the column of mercury is 1.74 cm. This is a large change compared to the original length of the column and shows why thermometers contain bulbs at the bottom instead of simple columns of liquid. The bulb at the bottom provides space for the liquid to expand without causing damage to the thermometer.
No7
A mercury thermometer still in use for meteorology has a bulb with a volume of 0.780 cm and a tube for the mercury to expand into of inside diameter 0.130 mm. (a) Neglecting the thermal expansion of the glass, what is the spacing between marks 1 "C apare (b) If the thermometer is made of ordinary glass (not a good idra), what is the spacing?
SOLUTION
To calculate the spacing between marks on the thermometer, we need to consider the volume change of the mercury as it expands or contracts with temperature changes. The formula we can use is:
ΔV = VαΔT
Where:
ΔV is the change in volume
V is the initial volume
α is the coefficient of volume expansion
ΔT is the change in temperature
a) Neglecting the thermal expansion of the glass, we can assume that the change in volume is solely due to the change in volume of the mercury. The coefficient of volume expansion for mercury is approximately 181.2 x 10^-6 °C^-1.
Let's assume a temperature change of 1 °C. We'll also assume that the initial volume of the mercury is equal to the volume of the bulb, which is 0.780 cm^3. Plugging these values into the formula, we have:
ΔV = (0.780 cm^3)(181.2 x 10^-6 °C^-1)(1 °C)
ΔV = 0.000141096 cm^3
The spacing between marks on the thermometer would be equal to this change in volume (ΔV).
b) If the thermometer is made of ordinary glass, we need to consider its thermal expansion as well. The coefficient of linear expansion for ordinary glass is approximately 9 x 10^-6 °C^-1.
Using the same temperature change of 1 °C and the same initial volume of the mercury, we can calculate the change in volume due to the expansion of both the mercury and the glass.
ΔV_mercury = (0.780 cm^3)(181.2 x 10^-6 °C^-1)(1 °C)
ΔV_mercury = 0.000141096 cm^3
ΔV_glass = (0.780 cm^3)(9 x 10^-6 °C^-1)(1 °C)
ΔV_glass = 0.00000702 cm^3
The total change in volume would be the sum of these two changes:
ΔV_total = ΔV_mercury + ΔV_glass
ΔV_total = 0.000141096 cm^3 + 0.00000702 cm^3
ΔV_total = 0.000148116 cm^3
Again, the spacing between marks on the thermometer would be equal to this total change in volume (ΔV_total).
No8
Even when shut down after a period of normal use. a large commercial nuclear reactor transfers thermal energy at the rate of 150 MW by the radioactive decay of fission products. This heat transfer causes a rapid increase in temperature if the cooling system fails.
SOLUTION
If the cooling system of a large commercial nuclear reactor fails, the heat transfer caused by the radioactive decay of fission products will lead to a rapid increase in temperature. Let's assume that the cooling system completely fails and the rate of thermal energy transferred by the radioactive decay remains constant at 150 MW.
To calculate the increase in temperature, we can use the equation:
Q = mcΔT
Where:
Q is the heat transferred (in joules)
m is the mass of the material (in kilograms)
c is the specific heat capacity of the material (in joules per kilogram per degree Celsius)
ΔT is the change in temperature (in degrees Celsius)
Since we know the rate of thermal energy transferred (Q) and the specific heat capacity (c), we need to find the mass of the material to determine the change in temperature.
Let's assume that the fission products have a specific heat capacity of 0.5 J/g°C, which is a reasonable value for solid or liquid materials. We need to convert MW to joules per second, and grams to kilograms, to ensure consistent units.
150 MW = 150 x 10^6 J/s
Let's assume a time period of 1 second for simplicity.
Q = mcΔT
150 x 10^6 J = (m/1000 kg)(0.5 J/g°C)(ΔT)
Rearranging the equation, we find:
ΔT = (150 x 10^6 J) / ((m/1000 kg)(0.5 J/g°C))
To calculate the change in temperature, we need to know the mass of the material. However, you haven't provided any information about the specific material in the reactor. The mass and composition of the fission products can vary in different reactors.
Without the specific mass of the material, we cannot calculate the change in temperature accurately. Additionally, there are other factors that come into play in such a scenario, like the heat transfer mechanism and the capacity of the reactor to dissipate heat without the cooling system.
No9
In some countries, liquid nitrogen is used on dairy tracks instead of mechanical refrigerators. A 3.00-hour delivery trip requires 200 L at liquid nitrogen, which has a density of 808 kg/m (a) Calculate the heat transfer necessary to evaporate this amours of liquid nitrogen and raise its temperature to 3.00 °C. (Use Cp and assume it is constant over the temperature range) This value is the amount of cooling the liquid nitrogen supplies. (b) What is this heat transfer rate in kilowatt/hours? (c) Compare the amount of cooling obtained hum melting an identical mass of U-C ice with that from evaporating the liquid nitrogen.
SOLUTION
To solve this problem, we'll need to use several equations and conversion factors. Let's start with part (a).
(a) Calculate the heat transfer necessary to evaporate the liquid nitrogen and raise its temperature to 3.00 °C.
First, we need to calculate the mass of the liquid nitrogen used:
mass = volume * density
mass = 200 L * 808 kg/m^3
mass = 161,600 kg
Next, we need to calculate the heat transfer using the equation:
Q = mass * Cp * ΔT
where Q is the heat transfer, mass is the mass of the liquid nitrogen, Cp is the specific heat capacity of liquid nitrogen, and ΔT is the change in temperature.
The specific heat capacity of liquid nitrogen, Cp, is approximately 2.08 kJ/kg·°C.
ΔT = 3.00 °C - (-196 °C) = 199 °C
Q = 161,600 kg * 2.08 kJ/kg·°C * 199 °C
Q = 66,907,456 kJ
So, the heat transfer necessary to evaporate the liquid nitrogen and raise its temperature to 3.00 °C is approximately 66,907,456 kJ.
(b) Convert the heat transfer rate to kilowatt/hours.
To convert from kilojoules to kilowatt/hours, we need to use the conversion factor:
1 kilojoule (kJ) = 2.78 x 10^-4 kilowatt/hour (kWh)
Heat transfer rate in kWh = Q * (1 kJ / 2.78 x 10^-4 kWh)
Heat transfer rate in kWh = 66,907,456 kJ * (1 / 2.78 x 10^-4)
Heat transfer rate in kWh ≈ 240,705,035 kWh
So, the heat transfer rate required is approximately 240,705,035 kWh.
(c) Compare the amount of cooling obtained from melting an identical mass of U-C ice with that from evaporating the liquid nitrogen.
To compare the amount of cooling, we need to calculate the heat transfer required to melt the same mass of U-C ice.
The specific heat capacity of U-C ice is approximately 2.09 kJ/kg·°C.
ΔT = 3.00 °C - (-20 °C) = 23 °C
Q = mass * Cp * ΔT
Q = 161,600 kg * 2.09 kJ/kg·°C * 23 °C
Q = 7,640,800 kJ
So, the heat transfer necessary to melt the same mass of U-C ice and raise its temperature to 3.00 °C is approximately 7,640,800 kJ.
Comparing the heat transfers:
Liquid nitrogen cooling: 66,907,456 kJ
U-C ice melting: 7,640,800 kJ
The amount of cooling obtained from evaporating the liquid nitrogen is significantly larger than the amount of cooling obtained from melting the same mass of U-C ice.
The coefficient of volume expansion for copper is approximately 0.000016 per °C. Assuming the radiator is initially filled at its capacity temperature of 100 °C and the overflow occurs at a lower temperature, we need to find the volume difference between these two temperatures.
Let's assume the lower temperature is 20 °C. Therefore, the temperature difference is 100 °C - 20 °C = 80 °C.
No10
A 0.800-kg iron cylinder at a temperature of 1.00 x 10 C is dropped into an insulated chest of 1.00 kg of ice at its melting point. What is the final temperature, and how much ice has melted?
SOLUTION
To solve this problem, we'll use the principles of heat transfer and the specific heat capacities of iron and ice.
First, let's calculate the heat transfer for the iron cylinder as it cools down and reaches the final temperature.
The specific heat capacity of iron (Cp_iron) is approximately 0.449 kJ/kg·°C.
The initial temperature of the iron cylinder (T_initial) is 1.00 x 10 °C, and the final temperature (T_final) is the same for both the iron cylinder and the melted ice.
The equation for heat transfer is:
Q_iron = mass_iron * Cp_iron * ΔT_iron
where Q_iron is the heat transfer for the iron cylinder, mass_iron is the mass of the iron cylinder, Cp_iron is the specific heat capacity of iron, and ΔT_iron is the change in temperature for the iron cylinder.
ΔT_iron = T_final - T_initial
ΔT_iron = T_final - (1.00 x 10 °C)
Q_iron = 0.800 kg * 0.449 kJ/kg·°C * ΔT_iron
Q_iron = 0.3592 kJ/°C * ΔT_iron
Next, let's calculate the heat transfer for the melted ice as it warms up and reaches the final temperature.
The specific heat capacity of ice (Cp_ice) is approximately 2.09 kJ/kg·°C.
The equation for heat transfer is:
Q_ice = mass_ice * Cp_ice * ΔT_ice
where Q_ice is the heat transfer for the melted ice, mass_ice is the mass of the melted ice, Cp_ice is the specific heat capacity of ice, and ΔT_ice is the change in temperature for the melted ice.
ΔT_ice = T_final - 0 °C
ΔT_ice = T_final
Q_ice = mass_ice * Cp_ice * ΔT_ice
Q_ice = 1.00 kg * 2.09 kJ/kg·°C * ΔT_ice
Q_ice = 2.09 kJ/°C * ΔT_ice
Since the system is insulated, the heat transfer for the iron cylinder is equal to the heat transfer for the melted ice:
Q_iron = Q_ice
0.3592 kJ/°C * ΔT_iron = 2.09 kJ/°C * ΔT_ice
Simplifying the equation:
0.3592 ΔT_iron = 2.09 ΔT_ice
Now, let's consider the mass of the ice that has melted.
We know that 1 kg of ice has melted. The heat required to melt the ice is given by the equation:
Q_melted_ice = mass_melted_ice * Latent_heat_of_fusion_ice
where Q_melted_ice is the heat transfer for the melted ice, mass_melted_ice is the mass of the melted ice, and Latent_heat_of_fusion_ice is the latent heat of fusion for ice.
The latent heat of fusion for ice (Latent_heat_of_fusion_ice) is approximately 334 kJ/kg.
Q_melted_ice = mass_melted_ice * Latent_heat_of_fusion_ice
We have already established that Q_iron = Q_ice, so Q_melted_ice = Q_ice.
mass_melted_ice * Latent_heat_of_fusion_ice = 2.09 kJ/°C * ΔT_ice
Now, we have two equations:
0.3592 ΔT_iron = 2.09 ΔT_ice
mass_melted_ice * 334 kJ/kg = 2.09 kJ/°C * ΔT_ice
From the first equation, we can solve for ΔT_iron:
ΔT_iron = (2.09/0.3592) ΔT_ice
Substituting this into the second equation:
mass_melted_ice * 334 kJ/kg = 2.09 kJ/°C * [(2.09/0.3592) ΔT_ice]
Simplifying the equation:
mass_melted_ice * 334 = (2.09 * 2.09 / 0.3592) ΔT_ice
Now, we have one equation with one unknown, ΔT_ice.
Let's substitute the values into the equation and solve for ΔT_ice:
ΔT_ice = (mass_melted_ice * 334) / ((2.09 * 2.09) / 0.3592)
Substituting the given mass_melted_ice = 1.00 kg:
ΔT_ice = (1.00 kg * 334) / ((2.09 * 2.09) / 0.3592)
ΔT_ice ≈ 6.25 °C
Now, we can substitute this value back into the first equation to find ΔT_iron:
ΔT_iron = (2.09/0.3592) ΔT_ice
ΔT_iron = (2.09/0.3592) * 6.25 °C
ΔT_iron ≈ 36.19 °C
To find the final temperature:
T_final = T_initial + ΔT_iron
T_final = 10 °C + 36.19 °C
T_final ≈ 46.19 °C
So, the final temperature is approximately 46.19 °C, and the amount of ice melted is 1.00 kg.
No11
A 30.0-g ice cube at its melting point is dropped into an aluminum calorimeter of mass 1000 g in equilibrium at 24.0°C with 300.0 g of an unknown liquid. The final temperature is 4.0 °C. What is the heat capacity of the liquid?
SOLUTION
To solve this problem, we can use the principle of conservation of energy.
First, let's find the heat gained by the calorimeter. The formula for heat is:
Q = mcΔT
Where:
Q = Heat gained (or lost)
m = mass
c = specific heat capacity
ΔT = change in temperature
Since the calorimeter is in equilibrium at 24.0°C, and the final temperature is 4.0°C, the change in temperature is:
ΔT_calorimeter = 4.0°C - 24.0°C = -20.0°C
The mass of the calorimeter is 1000 g, and the specific heat capacity of aluminum is 0.897 J/g°C. Plugging the values into the formula, we get:
Q_calorimeter = (1000 g)(0.897 J/g°C)(-20.0°C)
Q_calorimeter = -17,940 J
Next, let's find the heat lost by the ice cube. The mass of the ice cube is 30.0 g, and the heat of fusion of ice is 334 J/g (this is the amount of heat required to convert 1 g of ice at 0°C into 1 g of water at 0°C). Plugging the values into the formula, we get:
Q_ice = (30.0 g)(334 J/g)
Q_ice = 10,020 J
Since energy is conserved, the heat gained by the calorimeter is equal to the heat lost by the ice cube:
Q_calorimeter = Q_ice
-17,940 J = 10,020 J
Finally, let's find the heat capacity of the liquid. The mass of the liquid is 300.0 g. Plugging the values into the formula, we can solve for the specific heat capacity of the liquid:
Q_liquid = mcΔT
-17,940 J = (300.0 g)(c)(4.0°C - 24.0°C)
-17,940 J = (300.0 g)(c)(-20.0°C)
c = -17,940 J / (300.0 g * -20.0°C)
c ≈ 2.99 J/g°C
Therefore, the heat capacity of the liquid is approximately 2.99 J/g°C.
No12
(a) Calculate the rate of heat conduction through a double-paned window that has a 150-m area and is made of two panes of 0.800-cm-thick glass separated by a 100-cm air gap. The inside surface temperature is 15.0°C while that on the outside is -10.0 °C. (Hint There are identical temperature drops across the two glass panes. First find these and then the temperature drop across the air gap. This problem ignores the increased heat transfer in the air gap due to convection.) (h) Calculate the tate of heat conduction through a 1.60-cm-thick window of the same area and with the same temperatures. Compare your answer with that for part (a).
SOLUTION
a) To calculate the rate of heat conduction through the double-paned window, we can break it down into three parts: the heat conduction through the first glass pane, the heat conduction through the air gap, and the heat conduction through the second glass pane.
1. Heat conduction through the glass pane:
Using the formula for thermal conduction, the heat transfer rate through a material is given by:
Q = (k * A * ΔT) / d
where Q is the heat transfer rate, k is the thermal conductivity of the material, A is the area of the surface, ΔT is the temperature difference across the material, and d is the thickness of the material.
Since there are two glass panes, the temperature drop across each pane is half of the overall temperature difference.
ΔT_glass = (15.0 - (-10.0)) / 2 = 12.5 °C
The thermal conductivity of glass is typically around 0.96 W/(m·K).
k_glass = 0.96 W/(m·K)
The thickness of each glass pane is given as 0.800 cm, which is 0.008 m.
d_glass = 0.008 m
So the heat transfer rate through each glass pane is:
Q_glass = (k_glass * A * ΔT_glass) / d_glass
2. Heat conduction through the air gap:
The temperature difference across the air gap is the same as the overall temperature difference.
ΔT_air = 15.0 - (-10.0) = 25.0 °C
The thermal conductivity of air at atmospheric pressure is approximately 0.026 W/(m·K).
k_air = 0.026 W/(m·K)
The thickness of the air gap is given as 100 cm, which is 1.0 m.
d_air = 1.0 m
So the heat transfer rate through the air gap is:
Q_air = (k_air * A * ΔT_air) / d_air
3. Total heat conduction:
Since the heat transfer rates through each pane of glass and the air gap are in series, we can simply add them up to get the total heat transfer rate:
Q_total = 2 * Q_glass + Q_air
Substituting the values we have:
Q_total = 2 * ((0.96 W/(m·K)) * (150 m^2) * (12.5 °C)) / (0.008 m) + ((0.026 W/(m·K)) * (150 m^2) * (25.0 °C)) / (1.0 m)
Calculate to find the rate of heat conduction through the double-paned window.
(h) To calculate the rate of heat conduction through a single-pane window of the same area and with the same temperatures, we can use a similar approach:
1. Heat conduction through the glass pane:
We can use the same formula as before to calculate the heat transfer rate through the glass pane. However, in this case, we only have one glass pane rather than two.
So the heat transfer rate through the glass pane is:
Q_glass_single = (k_glass * A * ΔT_glass) / d_glass
2. Total heat conduction:
Since there is only one glass pane, we don't have to consider an air gap. We can simply use the heat transfer rate through the glass pane as the total heat transfer rate:
Q_total_single = Q_glass_single
Substituting the values we have:
Q_total_single = (0.96 W/(m·K)) * (150 m^2) * (12.5 °C) / (0.016 m)
Calculate to find the rate of heat conduction through the single-pane window.
To compare the answers for part (a) and part (h), divide the rate of heat conduction through the single-pane window (Q_total_single) by the rate of heat conduction through the double-paned window (Q_total).
No13
You have a Dewar flask (a laboratory vacuum flask) that has an open top and straight sides, as shown below. You fill it with water and put it into the freezer. It is effectively a perfect insulator, blocking all heat transfer, except an the top. After a time, ice forms on the surface of the water. The liquid water and the bottom surface of the ice, in contact with the liquid water, are at 0°C. The sup surface of the ice is at the same temperature as the air in the freezer, -18°C. Set the rate of heat flow through the ice equal to the rate of loss of heat of fusion as the water freezes. When the ice layer is 0.700 cm thick, find the rate in mis at which the ice is thickening.
SOLUTION
To find the rate at which the ice is thickening, we need to calculate the rate of heat flow through the ice and set it equal to the rate of heat of fusion.
First, let's determine the rate of heat flow through the ice. We can use the equation for heat conduction:
Q = (k * A * ΔT) / d
where Q is the rate of heat flow, k is the thermal conductivity of ice, A is the cross-sectional area of the ice, ΔT is the temperature difference across the ice, and d is the thickness of the ice.
For the ice layer to be thickening, the rate of heat flow through the ice should equal the rate of heat of fusion. The rate of heat of fusion can be calculated using the following equation:
Q_fusion = m * L_fusion
where Q_fusion is the rate of heat of fusion, m is the mass of the ice layer per unit time, and L_fusion is the latent heat of fusion of ice.
Since the ice layer is thickening, we can assume that the mass of the ice layer per unit time is equal to the density of ice multiplied by the rate at which the ice is thickening (dm/dt = ρ * d). The density of ice, ρ, is approximately 917 kg/m^3.
Now, let's write down the equation for the rate of heat flow through the ice and the rate of heat of fusion:
(k * A * ΔT) / d = ρ * d * L_fusion
We can rearrange this equation to solve for the rate at which the ice is thickening (dm/dt):
dm/dt = (k * A * ΔT) / (d^2 * L_fusion)
Now, let's plug in the given values:
k (thermal conductivity of ice) = 2.22 W/(m*K)
A (cross-sectional area of the ice) = 1 m^2
ΔT (temperature difference across the ice) = 0°C - (-18°C) = 18°C = 18 K
d (thickness of the ice) = 0.700 cm = 0.007 m
L_fusion (latent heat of fusion of ice) = 333,500 J/kg
Plugging in these values, we can calculate the rate at which the ice is thickening (dm/dt):
dm/dt = (2.22 W/(m*K) * 1 m^2 * 18 K) / ((0.007 m)^2 * 333,500 J/kg)
Simplifying the equation, we find:
dm/dt ≈ 1.8 x 10^-5 kg/s
Therefore, the rate at which the ice is thickening is approximately 1.8 x 10^-5 kg/s, or 18 micrograms/s.
No14
How many grams of coffee must evaporate from 350 g of coffee in a 100-g glass cup to coal the coffee and the cup from 95.0 °C to 45.0°C? Assume the coffee has the same thermal properties as water and that the average heat of vaporization is 2340 kJ/kg (560 kcal/g). Neglect heat losses through processes other than evaporation, as well as the change in mass of the coffee as it cools. Do the latter two assumptions cause your answer to be higher or lower than the true answer?
SOLUTION
To calculate the amount of coffee that must evaporate, we first need to determine the amount of heat that needs to be transferred.
The heat required to cool the coffee and the cup from 95.0°C to 45.0°C can be calculated using the equation:
Q = mcΔT
Where:
Q = heat energy
m = mass of coffee and cup
c = specific heat capacity (assumed to be the same as water, 4.18 J/g°C)
ΔT = change in temperature (95.0°C - 45.0°C = 50.0°C)
Plugging in the values:
Q = (350 + 100) g * 4.18 J/g°C * 50.0°C
Q = 88100 J
Next, we need to convert the heat energy to kilojoules (kJ):
Q = 88100 J * (1 kJ / 1000 J)
Q = 88.1 kJ
Since the heat required for evaporation is given as 2340 kJ/kg, we can calculate the mass of coffee that needs to evaporate using the equation:
Q = mL
Where:
Q = heat energy required for evaporation
m = mass of coffee that evaporates
L = specific latent heat of vaporization (2340 kJ/kg)
Plugging in the values:
88.1 kJ = m * 2340 kJ/kg
m = 0.0377 kg
Finally, we can convert the mass of coffee to grams:
m = 0.0377 kg * (1000 g / 1 kg)
m = 37.7 g
Therefore, 37.7 grams of coffee must evaporate from 350 g of coffee in a 100-g glass cup to cool the coffee and the cup from 95.0°C to 45.0°C.
The assumptions made in this calculation (neglecting heat losses and the change in mass of the coffee as it cools) would cause the answer to be higher than the true answer. In reality, some heat would be lost to the surroundings, and as the coffee cools, its mass would decrease slightly due to evaporation.
No15
(a) It is difficult to extinguish a fire on a crude oil tanker because each liter of crude oil releases 2.80 x 107 of energy when burned. To illustrate this difficulty, calculate the number of liners of water that must be expended to absorb the energy released by burning 1.00 L of crude oil, if the water's temperature rises from 200 °C to 100 °C. It bails, and the resulting steam's temperature rises to 300°C at constant pressure. (b) Discuss additional complications caused by the fact that crude all is less dense than water.
SOLUTION
To calculate the number of liters of water needed to absorb the energy released by burning 1.00 L of crude oil, we need to determine the amount of heat energy released by burning the oil.
The heat released by burning 1.00 L of crude oil can be calculated using the given energy release per liter:
Q = (2.80 x 10^7 J/L) * 1.00 L
Q = 2.80 x 10^7 J
Next, we need to calculate the amount of heat energy absorbed by the water as its temperature rises from 200 °C to 100 °C. This can be calculated using the equation:
Q = mcΔT
Where:
Q = heat energy
m = mass of water
c = specific heat capacity of water (assumed to be 4.18 J/g°C)
ΔT = change in temperature (200 °C - 100 °C = 100 °C)
We can assume the mass of water to be equal to 1.00 kg as 1 L of water is approximately equal to 1 kg.
Plugging in the values:
Q = (1.00 kg) * 4.18 J/g°C * 100 °C
Q = 41800 J
Since the heat energy absorbed by the water is equal to the heat energy released by burning the crude oil, we can calculate the number of liters of water needed:
41800 J = (4.18 x 10^4 J/L) * N L
N = 41800 J / (4.18 x 10^4 J/L)
N ≈ 1 L
Therefore, approximately 1 liter of water is needed to absorb the energy released by burning 1.00 L of crude oil.
(b) The fact that crude oil is less dense than water presents additional complications when trying to extinguish a fire on a crude oil tanker. Water being denser than oil causes the oil to float on top of the water. This makes it difficult for water to come into direct contact with the burning oil, reducing its effectiveness in cooling and extinguishing flames. The water tends to flow under the oil, evaporating and causing the oil to spread and create fire hazards.
Thus, fighting a fire on a crude oil tanker requires specialized firefighting techniques and equipment, such as foam or chemical extinguishers, to control and suppress the fire more effectively. These methods create a barrier that prevents the oil from igniting or re-igniting, allowing for better control and extinguishing of the fire.
No16
To help prevent frost damage, 4.00 kg of water at 0°C is sprayed onto a fruit tree. (a) How much hear transfer occurs as the water freezes? (b) How much would the temperature of the 200-kg tree decrease if this amount of heat transferred from the tree? Take the specific heat to be 3,35 kJ/kg "C, and assume that no phase change occurs in the tree.
SOLUTION
a) To find the heat transfer that occurs as the water freezes, we need to calculate the heat released during the phase change from liquid water to solid ice.
The heat released during the freezing process can be calculated using the equation:
Q = m * ΔHf
Where:
Q = heat transfer
m = mass of water
ΔHf = heat of fusion of water, which is 334 kJ/kg (assuming no phase change occurs in the tree)
Plugging in the values:
Q = (4.00 kg) * (334 kJ/kg)
Q = 1336 kJ
Therefore, 1336 kJ of heat transfer occurs as the water freezes.
(b) To calculate the decrease in temperature of the tree due to this heat transfer, we can use the equation:
Q = mcΔT
Where:
Q = heat transfer
m = mass of the tree (200 kg)
c = specific heat capacity of the tree (3.35 kJ/kg°C)
ΔT = change in temperature
Rearranging the equation, we can solve for ΔT:
ΔT = Q / (mc)
Plugging in the values:
ΔT = (1336 kJ) / (200 kg * 3.35 kJ/kg°C)
ΔT ≈ 2.00°C
Therefore, the temperature of the 200 kg tree would decrease by approximately 2.00°C due to the heat transferred from the freezing water.
No17
A 0.250 kg aluminum bowl holding 0.800 kg af soup at 25.0°C is placed in a freezer. What is the final temperature if 388 kJ of energy is transferred from the bowl and soup, assuming the soup's thermal properties are the same as that of water?
SOLUTION
We can use the formula:
Q = mcΔT
where Q is the amount of heat transferred, m is the mass of the substance, c is its specific heat capacity, and ΔT is the change in temperature.
Let's first find the initial temperature of the bowl and soup. We know that the soup is at 25.0°C, but we need to find the temperature of the bowl. Assuming that the bowl is in thermal equilibrium with the room, we can assume that its initial temperature is also 25.0°C.
Next, let's find the amount of heat transferred. We're given that 388 kJ of energy is transferred from the bowl and soup, so Q = -388,000 J (the negative sign indicates that heat is being removed from the bowl and soup).
The specific heat capacity of aluminum is 0.900 J/(g°C), and the specific heat capacity of water is 4.184 J/(g°C). We can calculate the total mass of the bowl and soup as 0.250 kg + 0.800 kg = 1.050 kg.
We can now use the formula to solve for the final temperature:
Q = mcΔT
-388,000 = (0.250 + 0.800) x (0.900 x ΔT + 4.184 x ΔT)
Simplifying this equation, we get:
-388,000 = 4.6944 ΔT
ΔT = -82.71°C
This negative value for ΔT means that the final temperature is 25.0°C - 82.71°C = -57.71°C. However, this is below the freezing point of water, so the soup would have frozen solid. Therefore, our assumption that the soup's thermal properties are the same as that of water may not be valid in this case.
No18
On a certain dry sunny day, a swimming pool's temperature would rise by 1.50°C if not for evaporation. What fraction of the water must evaporate to carry away precisely enough energy to keep the temperature constant?
SOLUTION
To keep the temperature constant, the amount of energy that needs to be carried away through evaporation must equal the amount of energy gained by the water from the surroundings.
The amount of energy gained by the water can be calculated using the specific heat capacity formula:
Energy gained = mass * specific heat capacity * change in temperature
Let's assume the mass of the water in the swimming pool is 1 kg (you can choose any other value, the final answer will be the same).
The specific heat capacity of water is approximately 4,186 J/kg°C.
The change in temperature is 1.50°C.
Energy gained = 1 kg * 4,186 J/kg°C * 1.50°C
Energy gained = 6,279 J
Now, we need to find the fraction of the water that needs to evaporate in order to carry away this energy.
The heat of vaporization of water is approximately 2,257,000 J/kg.
Let's assume the fraction of water that evaporates is x.
The energy carried away through evaporation = mass evaporated * heat of vaporization
Since the mass evaporated is a fraction of the initial mass (1 kg), we can say:
mass evaporated = x * (initial mass)
Energy carried away = x * (initial mass) * heat of vaporization
We need the energy carried away to be equal to the energy gained, so we set up the equation:
x * (initial mass) * heat of vaporization = energy gained
Plugging in the values we found:
x * 1 kg * 2,257,000 J/kg = 6,279 J
Simplifying the equation:
x * 2,257,000 J = 6,279 J
x = 6,279 J / 2,257,000 J
x ≈ 0.00278
Therefore, approximately 0.278% of the water in the swimming pool must evaporate to carry away enough energy to keep the temperature constant.
No19
A bag containing 0°C ice is much more effective in absorbing energy than one containing the same amount of 0°C water (a) How much heat transfer is necessary to raise the temperature of 0.800 kg of water from 0°C to 30.0°C? (b) How much heat transfer is required to first melt 0.800 kg of 0°C ice and then raise its temperature? (C) Explain how your answer supports the contention that the ice is more effective.
(a) To calculate the heat transfer required to raise the temperature of 0.800 kg of water from 0°C to 30.0°C, we can use the formula:
Heat transfer = mass * specific heat capacity * change in temperature
The specific heat capacity of water is approximately 4,186 J/kg°C.
Mass of water = 0.800 kg
Change in temperature = 30.0°C - 0°C = 30.0°C
Heat transfer = 0.800 kg * 4,186 J/kg°C * 30.0°C
Heat transfer = 100,368 J
Therefore, it takes 100,368 Joules of heat transfer to raise the temperature of 0.800 kg of water from 0°C to 30.0°C.
(b) To calculate the heat transfer required to melt 0.800 kg of 0°C ice and then raise its temperature, we need to consider two separate steps: heat transfer for melting the ice and heat transfer for raising the temperature of the resulting water.
Step 1: Heat transfer for melting the ice
The heat of fusion of ice is approximately 334,000 J/kg.
Heat transfer for melting ice = mass of ice * heat of fusion
mass of ice = 0.800 kg
Heat transfer for melting ice = 0.800 kg * 334,000 J/kg
Heat transfer for melting ice = 267,200 J
Step 2: Heat transfer for raising the temperature of water from 0°C to 30.0°C (water's specific heat capacity)
Heat transfer = mass of water * specific heat capacity * change in temperature
Mass of water = 0.800 kg
Change in temperature = 30.0°C - 0°C = 30.0°C
Heat transfer for raising temperature = 0.800 kg * 4,186 J/kg°C * 30.0°C
Heat transfer for raising temperature = 100,368 J
Total heat transfer = Heat transfer for melting ice + Heat transfer for raising temperature
Total heat transfer = 267,200 J + 100,368 J
Total heat transfer = 367,568 J
Therefore, it takes 367,568 Joules of heat transfer to first melt 0.800 kg of 0°C ice and then raise its temperature.
(c) The higher amount of heat transfer required to melt the ice and raise its temperature compared to just raising the temperature of the water shows that the ice is more effective in absorbing energy. This is because the process of melting ice requires a significant amount of energy due to the heat of fusion, which is much higher than the specific heat capacity of water. The energy required to melt the ice comes from the surroundings and is absorbed by the ice, resulting in a larger heat transfer compared to just raising the temperature of water. Thus, the ice is more efficient in absorbing energy compared to water.
No20
How much heat transfer (in kilocalories) is required to thaw a 0.450-kg package at frozen vegetables originally at 0°C if their heat of fusion is the same as that of water?
SOLUTION
The heat of fusion of water is approximately 334 kiloJoules per kilogram (kJ/kg).
To convert this to kilocalories, we divide by 4.184 (1 kilocalorie = 4.184 kilojoules).
Heat of fusion of water in kilocalories = 334 kJ/kg / 4.184
Heat of fusion of water in kilocalories = 79.798 kcal/kg
Mass of frozen vegetables = 0.450 kg
Heat transfer required to thaw the package = Mass of frozen vegetables * Heat of fusion of water in kilocalories
Heat transfer required to thaw the package = 0.450 kg * 79.798 kcal/kg
Heat transfer required to thaw the package = 35.909 kcal
Therefore, it takes 35.909 kilocalories of heat transfer to thaw a 0.450-kg package of frozen vegetables originally at 0°C, assuming their heat of fusion is the same as that of water.
Overall, this blog post provides a comprehensive overview of temperature, heat transfer, and related concepts, serving as a useful resource for readers looking to deepen their understanding of these topics.
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