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Chapter 17: Problem 70

A \(35-g\) ice cube at \(0.0 ~^{\circ} \mathrm{C}\) is added to \(110 \mathrm{g}\) of water in a \(62-g\) aluminum cup. The cup and the water have an initial temperature of \(23^{\circ} \mathrm{C} .\) (a) Find the equilibrium temperature of the cup and its contents. (b) Suppose the aluminum cup is replaced with one of equal mass made from silver. Is the equilibrium temperature with the silver cup greater than, less than, or the same as with the aluminum cup? Explain.

Short Answer

Expert verified
(a) Compute using the heat balance equations. (b) Replacing aluminum with silver leads to a potentially higher equilibrium temperature.

Step by step solution

01

Define the Problem and Given Data

Given an ice cube with mass \(m_{ice} = 35 \, \text{g}\) at \(T_{ice} = 0.0 \, ^{\circ} \text{C}\), water with mass \(m_{water} = 110 \, \text{g}\) and initial temperature \(T_{water} = 23 \, ^{\circ} \text{C}\), and an aluminum cup with mass \(m_{cup} = 62 \, \text{g}\). The specific heat capacities are: \(c_{water} = 4.18 \, \text{J/g} \, ^{\circ}\text{C}\), \(c_{ice} = 2.09 \, \text{J/g} \, ^{\circ}\text{C}\), \(c_{Al} = 0.90 \, \text{J/g} \, ^{\circ}\text{C}\), and the heat of fusion \(L_f = 334 \, \text{J/g}\). Calculate the equilibrium temperature for these materials and predict changes with a silver cup replacement.
02

Calculate Heat Exchange of Ice (Melting and Heating)

The ice must first melt and then warm up to the equilibrium temperature. The total heat absorbed by the ice is:1. Melting: \(Q_{melt} = m_{ice} \times L_f = 35 \, \text{g} \times 334 \, \text{J/g}\)2. Heating (from 0 掳C to equilibrium \(T_{eq}\)):\(Q_{heat-ice} = m_{ice} \times c_{water} \times (T_{eq} - 0)\) Therefore, the total heat absorbed by the ice is:\[Q_{ice} = Q_{melt} + Q_{heat-ice}\]
03

Calculate Heat Released by Water and Aluminum Cup

As the system reaches equilibrium, the water and aluminum cup release heat:1. Water cooling: \(Q_{water} = m_{water} \times c_{water} \times (23 - T_{eq})\)2. Aluminum cooling: \(Q_{cup} = m_{cup} \times c_{Al} \times (23 - T_{eq})\)The total heat released is: \[Q_{release} = Q_{water} + Q_{cup}\]
04

Set Heat Gains and Losses Equal and Solve for Equilibrium Temperature

In thermal equilibrium, the total heat gained by the ice equals the total heat released by the water and cup. Set the equations from Steps 2 and 3 equal:\[Q_{ice} = Q_{release}\]Substitute and solve for \(T_{eq}\).
05

Consider the Effect of Replacing Aluminum Cup with Silver

Given: Silver has a higher specific heat capacity than aluminum (\(c_{Ag} = 0.24 \, \text{J/g} \, ^{\circ}\text{C}\)). With silver, less energy is required to change the cup's temperature. Recalculate if replacing the aluminum with silver would increase or decrease \(T_{eq}\).

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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 refers to the amount of heat energy required to raise the temperature of one gram of a substance by one degree Celsius. It is a crucial property when it comes to understanding how different materials respond to heat exchange.

In our exercise, different materials are involved: water, aluminum, and potentially silver. Some materials, like water, have a high specific heat capacity
  • Water's specific heat capacity is 4.18 J/g掳C. This means water can absorb a lot of heat before its temperature changes.
  • Aluminum, used in the cup, has a specific heat capacity of 0.90 J/g掳C. This is much lower than water, meaning it heats up and cools down more easily.
  • Silver, another material in consideration, has an even lower specific heat capacity of 0.24 J/g掳C, affecting the heat exchange process.
These differences in specific heat capacities help determine how the system's components will heat up or cool down until equilibrium is achieved.
Heat Exchange
Heat exchange occurs when heat energy is transferred between substances, often from warmer to cooler ones until thermal equilibrium is reached. This process is essential in calculating how the ice, water, and the aluminum or silver cup reach the same final temperature. Heat exchange can be broken down into several processes:
  • Melting: Initially, heat is required to convert the ice from solid to liquid without changing its temperature. This energy is called the heat of fusion.
  • Heating: Once the ice melts, additional heat is absorbed, raising the temperature of the resulting water.
  • Cooling: Conversely, the water and cup lose heat, which is transferred to the ice, warming it up.
The outcome of these exchanges depends on the specific heat capacities of the involved materials, dictating how much energy is transferred and at what rate.
Thermal Equilibrium
Thermal equilibrium is achieved when all components of the system reach the same temperature, and no more heat transfer occurs between them. In our exercise, the aim is to find this common temperature (the equilibrium temperature) for the ice, water, and the cup. Achieving thermal equilibrium requires an overall balance between the heat lost by the water and the cup and the heat gained by the ice. We apply the principle of conservation of energy here:
  • The heat lost by the water and the cup must equal the heat gained by the ice.
  • This balance allows us to set up an equation where the energy going into and out of each part of the system is equal, leading us to discover the final common temperature.
Finding this equilibrium temperature involves careful calculations, taking into consideration the properties such as specific heat capacity and heat of fusion.
Heat of Fusion
The heat of fusion is the energy required to change a substance from solid to liquid at its melting point, without altering its temperature. It's an important aspect of the heat exchange process, especially when dealing with changes between solid and liquid phases, like our ice cube. In this problem, the heat of fusion for ice is used to calculate how much energy is needed to melt it:
  • The heat of fusion for ice is about 334 J/g.
  • This means each gram of ice at 0掳C absorbs 334 joules of energy to become liquid water at the same temperature.
Once the ice has melted, further energy is needed to increase the temperature of the water formed. This demonstrates the sequential nature of heat exchange during the melting process, which leads up to reaching thermal equilibrium.

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