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Chapter 18: Problem 41

(a) Two \(50 \mathrm{~g}\) ice cubes are dropped into \(200 \mathrm{~g}\) of water in a thermally insulated container. If the water is initially at \(25^{\circ} \mathrm{C},\) and the ice comes directly from a freezer at \(-15^{\circ} \mathrm{C},\) what is the final temperature at thermal equilibrium? (b) What is the final temperature if only one ice cube is used?

Short Answer

Expert verified
(a) Final temperature: 0掳C (all ice melts); (b) Final temperature: less than 0掳C (not all ice melts).

Step by step solution

01

Understand the Problem

We are given two scenarios: (a) adding two ice cubes to water and (b) adding one ice cube to water. The water and ice need to reach thermal equilibrium without any heat loss to the environment as the container is insulated. We'll calculate the final temperature in each scenario.
02

List Given Data

The mass of water (\(m_w\)) is 200 g, the mass of each ice cube (\(m_i\)) is 50 g. The initial temperature of the water is \(T_w = 25^{\circ} \text{C}\) and the initial temperature of the ice is \(T_i = -15^{\circ} \text{C}\). Specific heat capacities: \(c_w = 4.18 \text{ J/g掳C}, \, c_i = 2.09 \text{ J/g掳C}\), and the latent heat of fusion for ice is \(L_f = 334 \text{ J/g}\).
03

Calculate Heat Required to Warm Ice to 0掳C

Use the formula \( q_1 = m_i \, c_i \, (0 - T_i) \).For two cubes, \( q_{1a} = 2 \, \times \, 50 \, \times \, 2.09 \, \times \, 15 \) J.For one cube, \( q_{1b} = 50 \, \times \, 2.09 \, \times \, 15 \) J.
04

Calculate Heat Required for Melting Ice

Use the formula \( q_2 = m_i \, L_f \).For two cubes, \( q_{2a} = 2 \, \times \, 50 \, \times \, 334 \) J.For one cube, \( q_{2b} = 50 \, \times \, 334 \) J.
05

Calculate Total Heat Required for Ice to Reach 0掳C Liquid

Add the values from the previous steps: \( q_{icea} = q_{1a} + q_{2a} \) and \( q_{iceb} = q_{1b} + q_{2b} \).Calculate both values.
06

Calculate Heat Loss from Water to Reach Equilibrium

The heat lost by water is \( q_{water} = m_w \, c_w \, (T_f - T_w) \), where \( T_f \) is the final temperature. Set \( q_{ice} = q_{water} \) and solve for \( T_f \) in each scenario.
07

Solve for Final Temperature for Two Ice Cubes

Set up the equation \( q_{icea} = m_w \, c_w \, (T_f - T_w) \) and solve for \( T_f \). This involves substituting the calculated values of \( q_{icea} \) and the known values for the heat capacity and mass of water.
08

Solve for Final Temperature for One Ice Cube

Set up the equation \( q_{iceb} = m_w \, c_w \, (T_f - T_w) \) similar to the previous step. Substitute the calculated \( q_{iceb} \) and solve for the final temperature \( T_f \).

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Key Concepts

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

Specific Heat Capacity
Specific heat capacity is an important concept when studying thermal equilibrium and heat transfer. It represents the amount of heat energy required to change the temperature of a unit mass of a substance by one degree Celsius. In this exercise, we are dealing with two different specific heat capacities: the specific heat capacity of water and ice.
- For water, the specific heat capacity is high, around 4.18 J/g掳C. This means water requires a relatively large amount of energy to change its temperature.
- Ice, in contrast, has a lower specific heat capacity of around 2.09 J/g掳C, one of the reasons it warms up faster than water.
Knowing these values is crucial for calculating how much heat is absorbed by the melting ice and how much is lost from the water until both reach the same final temperature.
Latent Heat of Fusion
The latent heat of fusion is the amount of heat required to change a substance from a solid to a liquid without changing its temperature. In our problem, this refers to the energy needed to melt the ice.
- Ice has a latent heat of fusion of 334 J/g. This energy is absorbed from the surrounding environment, which in this insulated scenario, is the warmer water.
When calculating the energy required to transition ice at 0掳C to water at 0掳C, it is crucial to multiply the mass of the ice by its latent heat of fusion. Hence, it represents a crucial step in understanding how energy is conserved and transferred in thermodynamic processes.
Calorimetry
Calorimetry is the study of measuring heat transfer in physical and chemical processes. In this exercise, it is used to analyze how the heat lost by water is equal to the heat gained by ice, ensuring that energy is conserved in the insulated system.
- The container mentioned is thermally insulated, implying no heat can enter or escape, vital for precise calorimetric calculations.
This concept involves setting the heat lost by the water equal to the total heat gained by the ice. Ultimately, calorimetry aids in calculating the final temperature when all components reach equilibrium. It underlines the principle that in a closed system like this, energy is conserved, and we can solve for unknown variables like final temperature.
Heat Transfer
Heat transfer is the process whereby thermal energy moves from a hotter object to a cooler one until thermal equilibrium is reached. This problem demonstrates heat transfer using both sensible and latent heat.
- *Sensible heat transfer* involves a temperature change in the water and ice without a phase change.
- *Latent heat transfer* occurs when the ice absorbs energy to transition from solid to liquid, without changing its temperature.
Understanding heat transfer mechanisms allows us to determine how energy is exchanged and how equilibrium is achieved. In this specific exercise, it is about how efficiently heat transfers from warm water to cold ice and how it affects their temperatures until they balance out.

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Most popular questions from this chapter

A thermometer of mass \(0.0550 \mathrm{~kg}\) and of specific heat \(0.837 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\) reads \(15.0^{\circ} \mathrm{C} .\) It is then completely immersed in \(0.300 \mathrm{~kg}\) of water, and it comes to the same final temperature as the water. If the thermometer then reads \(44.4^{\circ} \mathrm{C},\) what was the temperature of the water before insertion of the thermometer?

A sample of gas expands from an initial pressure and volume of \(10 \mathrm{~Pa}\) and \(1.0 \mathrm{~m}^{3}\) to a final volume of \(2.0 \mathrm{~m}^{3} .\) During the expansion, the pressure and volume are related by the equation \(p=a V^{2},\) where \(a=10 \mathrm{~N} / \mathrm{m}^{8} .\) Determine the work done by the gas during this expansion.

Liquid water coats an active (growing) icicle and extends up a short, narrow tube along the central axis (Fig. 18-55). Because the water-ice interface must have a temperature of \(0^{\circ} \mathrm{C}\), the water in the tube cannot lose energy through the sides of the icicle or down through the tip because there is no temperature change in those directions. It can lose energy and freeze only by sending energy up (through distance \(L\) ) to the top of the icicle, where the temperature \(T_{r}\) can be below \(0^{\circ} \mathrm{C}\). Take \(L=0.12 \mathrm{~m}\) and \(T_{r}=-5^{\circ} \mathrm{C}\). Assume that the central tube and the upward conduction path both have cross-sectional area \(A .\) In terms of \(A,\) what rate is (a) energy conducted upward and (b) mass converted from liquid to ice at the top of the central tube? (c) At what rate does the top of the tube move downward because of water freezing there? The thermal conductivity of ice is \(0.400 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K},\) and the density of liquid water is \(1000 \mathrm{~kg} / \mathrm{m}^{3}\).

The giant hornet Vespa mandarinia japonica preys on Japanese bees. However, if one of the hornets attempts to invade a beehive, several hundred of the bees quickly form a compact ball around the hornet to stop it. They don't sting, bite, crush, or suffocate it. Rather they overheat it by quickly raising their body temperatures from the normal \(35^{\circ} \mathrm{C}\) to \(47^{\circ} \mathrm{C}\) or \(48^{\circ} \mathrm{C}\) which is lethal to the hornet but not to the bees (Fig. 18-44). Assume the following: 500 bees form a ball of radius \(R=2.0 \mathrm{~cm}\) for a time \(t=20\) min, the primary loss of energy by the ball is by thermal radiation, the ball's surface has emissivity \(\varepsilon=0.80\), and the ball has a uniform temperature. On average, how much additional energy must each bee produce during the 20 min to maintain \(47^{\circ} \mathrm{C} ?\)

Consider the liquid in a barometer whose coefficient of volume expansion is \(6.6 \times 10^{-4} / \mathrm{C}^{\circ} .\) Find the relative change in the liquid's height if the temperature changes by \(12 \mathrm{C}^{\circ}\) while the pressure remains constant. Neglect the expansion of the glass tube.
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