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Chapter 18: Problem 68

For a class demonstration, your physics instructor pours \(1.00 \mathrm{~kg}\) of steam at \(100.0^{\circ} \mathrm{C}\) over \(4.00 \mathrm{~kg}\) of ice at \(0.00^{\circ} \mathrm{C}\) and allows the system to reach equilibrium. He is then going to measure the temperature of the system. While the system reaches equilibrium, you are given the latent heats of ice and steam and the specific heat of water: \(L_{\text {ice }}=3.33 \cdot 10^{5} \mathrm{~J} / \mathrm{kg}\), \(L_{\text {steam }}=2.26 \cdot 10^{6} \mathrm{~J} / \mathrm{kg}, c_{\text {water }}=4186 \mathrm{~J} /\left(\mathrm{kg}^{\circ} \mathrm{C}\right) .\) You are asked to calculate the final equilibrium temperature of the system. What value do you find?

Short Answer

Expert verified
Based on the given problem where steam at 100掳C is poured over ice at 0掳C, calculate the final equilibrium temperature of the system.

Step by step solution

01

Calculate the heat lost by steam

To find the heat lost by the steam, we first need to calculate the heat required to condense the steam into water. This can be done using the latent heat of steam: \(Q_1 = m_{steam} \cdot L_{steam}\) Now, plug in the values: \(Q_1 = 1.00 \mathrm{~kg} \cdot 2.26 \cdot 10^{6} \mathrm{~J} / \mathrm{kg} = 2.26 \cdot 10^{6} \mathrm{~J}\)
02

Calculate the heat gained by ice

To find the heat gained by the ice, we first need to calculate the heat required to melt the ice into water. This can be done using the latent heat of ice: \(Q_2 = m_{ice} \cdot L_{ice}\) Now, plug in the values: \(Q_2 = 4.00 \mathrm{~kg} \cdot 3.33 \cdot 10^{5} \mathrm{~J} / \mathrm{kg} = 1.33 \cdot 10^{6} \mathrm{~J}\)
03

Compare heat transfer values

Compare the values of \(Q_1\) and \(Q_2\): If \(Q_1 > Q_2\), then all ice melts and the final state is water only. In this case, we will then calculate the equilibrium temperature assuming all ice has melted. If \(Q_1 < Q_2\), then all steam condenses and the final state is a mixture of water and ice. In this case, the equilibrium temperature will be \(0^{\circ}C\). Here, \(Q_1 = 2.26 \cdot 10^{6} \mathrm{~J} > Q_2 = 1.33 \cdot 10^{6} \mathrm{~J}\), which means all the ice melts and we have only water in the final state.
04

Calculate the final equilibrium temperature

Since all ice has melted, we will now calculate the heat required to raise the temperature of the water formed from ice: \(Q_3 = m_{ice} \cdot c_{water} \cdot \Delta T_{ice}\) And calculate the heat required to lower the temperature of the water formed from steam: \(Q_4 = m_{steam} \cdot c_{water} \cdot \Delta T_{steam}\) We set both heat transfer values to be equal, as the heat lost by the steam-water must be gained by the ice-water: \(Q_3 = Q_4 \Rightarrow m_{ice} \cdot c_{water} \cdot \Delta T_{ice} = m_{steam} \cdot c_{water} \cdot \Delta T_{steam}\) The equilibrium temperature can then be calculated by solving for the temperature difference: \(\Delta T_{eq} = \frac{m_{steam} \cdot \Delta T_{steam}}{m_{ice}}\) We know the initial temperatures of steam and ice, so we can find the temperature differences: \(\Delta T_{steam} = 100^{\circ} \mathrm{C} - T_{eq}\) \(\Delta T_{ice} = T_{eq} - 0^{\circ} \mathrm{C}\) Substituting the initial temperature values and masses of steam and ice, we find the equilibrium temperature: \(T_{eq} = \frac{1.00 \mathrm{~kg}\cdot(100^{\circ} \mathrm{C} - T_{eq})}{4.00 \mathrm{~kg}}\) Solve for \(T_{eq}\): \(T_{eq} = \frac{100}{5} = 20 ^{\circ} \mathrm{C}\) The final equilibrium temperature of the system is \(20^{\circ} \mathrm{C}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Latent Heat
Latent heat is a significant concept in thermodynamics, relating to phase changes. When a substance changes its state, such as from solid to liquid or liquid to gas, latent heat is the energy absorbed or released without a change in temperature. This energy is referred to as "latent" because it is not detected by a thermometer since it doesn't change the substance's temperature during the phase change process.
In the classroom demonstration given in the problem, latent heat plays a crucial role as the steam condenses to water, and the ice melts into water. Each phase change requires or releases a specific amount of energy per kilogram of the substance, referred to as the latent heat of fusion for melting ice, and the latent heat of vaporization for condensing steam. In the given scenario:
  • Ice's latent heat of fusion: 3.33 x 10鈦� J/kg
  • Steam's latent heat of vaporization: 2.26 x 10鈦� J/kg
These values help determine how much energy is needed to fully melt the ice or fully condense the steam, illustrating the concept of energy conservation when systems reach thermal equilibrium.
Specific Heat Capacity
Specific heat capacity is another fundamental aspect of thermodynamics. It describes the amount of heat required to change the temperature of one kilogram of a substance by one degree Celsius. Different materials have different specific heat capacities, which defines how they respond to heat transfer.
In the problem, after the phase changes, the system's heat dynamics involve warming or cooling water, formed from ice and steam. The specific heat capacity for water is given as 4186 J/(kg掳C). This value allows us to calculate the energy exchanges needed to adjust the water temperature until equilibrium.
  • To heat or cool water by one degree, 4186 J would be needed per kilogram.
The large specific heat of water means it can absorb or release a substantial amount of heat without much change in temperature, playing a crucial role in stabilizing temperatures in such thermodynamic systems.
Equilibrium Temperature
The equilibrium temperature is the final stable temperature reached by a system where no more heat flows between the components. In any closed system exchange involving heat, the principle of conservation of energy applies, meaning the heat lost by one part of the system must equal the heat gained by another.
In the model problem, once all the steam condenses to liquid and the ice melts entirely to water, the system is assumed to reach thermal equilibrium. The calculations balance the heat absorbed by the melting ice and the heat lost by the condensing steam until no net heat exchange occurs, leading to temperature stability.
The process involves:
  • Determining heat transfer values of each phase change.
  • Setting the heat lost by the steam equal to the heat gained by the ice-water to find the equilibrium temperature.
Using the specific heat capacity of water and energy conservation principles, we solve for the system's final temperature, revealing the equilibrium at 20掳C in this specific exercise scenario.

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Most popular questions from this chapter

How would the rate of heat transfer between a thermal reservoir at a higher temperature and one at a lower temperature differ if the reservoirs were in contact with a 10 -cm-long glass rod instead of a 10 -m-long aluminum rod having an identical cross-sectional area?

A cryogenic storage container holds liquid helium, which boils at \(4.2 \mathrm{~K}\). Suppose a student painted the outer shell of the container black, turning it into a pseudoblackbody, and that the shell has an effective area of \(0.50 \mathrm{~m}^{2}\) and is at \(3.0 \cdot 10^{2} \mathrm{~K}\). a) Determine the rate of heat loss due to radiation. b) What is the rate at which the volume of the liquid helium in the container decreases as a result of boiling off? The latent heat of vaporization of liquid helium is \(20.9 \mathrm{~kJ} / \mathrm{kg} .\) The density of liquid helium is \(0.125 \mathrm{~kg} / \mathrm{L}\).

A \(100 .\) mm by \(100 .\) mm by 5.00 mm block of ice at \(0^{\circ} \mathrm{C}\) is placed on its flat face on a 10.0 -mm-thick metal disk that covers a pot of boiling water at normal atmospheric pressure. The time needed for the entire ice block to melt is measured to be \(0.400 \mathrm{~s} .\) The density of ice is \(920 . \mathrm{kg} / \mathrm{m}^{3} .\) Use the data in Table 18.3 to determine the metal the disk is most likely made of

You were lost while hiking outside wearing only a bathing suit. a) Calculate the power radiated from your body, assuming that your body's surface area is about \(2.00 \mathrm{~m}^{2}\) and your skin temperature is about \(33.0^{\circ} \mathrm{C} .\) Also, assume that your body has an emissivity of 1.00 . b) Calculate the net radiated power from your body when you were inside a shelter at \(20.0^{\circ} \mathrm{C}\). c) Calculate the net radiated power from your body when your skin temperature dropped to \(27.0^{\circ} \mathrm{C}\).

Arthur Clarke wrote an interesting short story called "A Slight Case of Sunstroke." Disgruntled football fans came to the stadium one day equipped with mirrors and were ready to barbecue the referee if he favored one team over the other. Imagine the referee to be a cylinder filled with water of mass \(60.0 \mathrm{~kg}\) at \(35.0^{\circ} \mathrm{C}\). Also imagine that this cylinder absorbs all the light reflected on it from 50,000 mirrors. If the heat capacity of water is \(4.20 \cdot 10^{3} \mathrm{~J} /\left(\mathrm{kg}^{\circ} \mathrm{C}\right),\) how long will it take to raise the temperature of the water to \(100 .{ }^{\circ} \mathrm{C}\) ? Assume that the Sun gives out \(1.00 \cdot 10^{3} \mathrm{~W} / \mathrm{m}^{2},\) the dimensions of each mirror are \(25.0 \mathrm{~cm}\) by \(25.0 \mathrm{~cm},\) and the mirrors are held at an angle of \(45.0^{\circ}\)
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