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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) The final temperature with two ice cubes is approximately 10.4掳C. (b) The final temperature with one ice cube is approximately 3掳C.

Step by step solution

01

Identify the Heat Transfer Components

First, we need to account for the heat transfer between the water and the ice cubes. The system comprises warm water and ice initially below the freezing point. Heat will be transferred to the ice to first raise its temperature to 0掳C, melt the ice, and then raise the temperature of the melted water.
02

Calculate Heat Required to Warm Ice to 0掳C

Each ice cube has a mass of 50 g (or 0.050 kg) and is initially at -15掳C. The heat required to warm each ice cube to 0掳C is calculated using the specific heat capacity of ice, which is approximately 2090 J/(kg路掳C). For one ice cube, the heat required is:\[Q_1 = (0.050 ext{ kg}) imes (2090 ext{ J/(kg掳C)}) imes (0 - (-15)掳C) = 1567.5 ext{ J}.\]
03

Calculate Heat Required to Melt Ice

The heat required to melt ice at 0掳C is calculated with the latent heat of fusion for ice, approximately 334,000 J/kg. For one ice cube:\[Q_2 = (0.050 ext{ kg}) imes (334,000 ext{ J/kg}) = 16700 ext{ J}.\] Thus, the total heat to bring one ice cube to water at 0掳C is \(Q_{1} + Q_{2} = 1567.5 + 16700 = 18267.5 ext{ J}.\)
04

Calculate Total Heat Absorbed by Two Ice Cubes

Since there are two ice cubes, the total heat absorbed is twice the amount calculated for one:\[Q_{ ext{total ice}} = 2 imes 18267.5 ext{ J} = 36535 ext{ J}.\]
05

Calculate Heat Lost by Water

The water's initial temperature is 25掳C, and we assume a final temperature of \(T_f\). The heat lost by the water is calculated using the specific heat capacity of water, approximately 4186 J/(kg路掳C). With a 200 g mass:\[Q_{ ext{water}} = (0.200 ext{ kg}) imes (4186 ext{ J/(kg掳C)}) imes (25 - T_f)掳C.\]
06

Set Heat Gained by Ice Equal to Heat Lost by Water

Since the system is thermally insulated, heat gained by the ice equals the heat lost by the water:\[36535 ext{ J} = (0.200 ext{ kg}) imes (4186 ext{ J/(kg掳C)}) imes (25 - T_f)掳C.\] Solve for \(T_f\).
07

Solve for Final Temperature with Two Ice Cubes

Solve the equation from Step 6:\[36535 = 837.2 imes (25 - T_f)\]\[T_f = 25 - \frac{36535}{837.2} \approx 10.4掳C.\]
08

Calculate for One Ice Cube

Repeat Steps 2 to 7, but use the heat calculated for one ice cube (18267.5 J). Set up the equation:\[18267.5 = 837.2 imes (25 - T_f)\]\[T_f = 25 - \frac{18267.5}{837.2} \approx 3掳C.\]

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Key Concepts

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

Heat Transfer
Heat transfer is the process of energy moving from one material or substance to another. In this exercise, we look at the thermal interaction between warm water and ice cubes placed within an insulated container.
During this process, heat transfers from the water to the ice cubes, causing a change in both their temperatures.
  • Warm water loses heat.
  • Ice cubes gain heat.
  • The system eventually reaches thermal equilibrium, where the temperatures of the water and the melted ice become equal.
This heat transfer continues until the energy given off by the water is exactly equal to the energy absorbed by the ice cubes. A perfect scenario is assured by using a thermally insulated container, which minimizes energy loss to the surroundings, allowing for clean calculations.
Specific Heat Capacity
Specific heat capacity describes how much heat energy is required to raise the temperature of a certain amount of a substance by one degree Celsius. It is a critical factor in solving the given problem as it helps determine how much energy is needed to change the temperature.
In the example, both ice and water have their unique specific heat capacities:
  • Ice has a specific heat capacity of approximately 2090 J/(kg路掳C).
  • Water's specific heat capacity is higher, around 4186 J/(kg路掳C).
The difference in these values shows that water needs more energy than ice per kilogram per degree to change its temperature. It is why specific heat capacity heavily influences the energy calculations for both warming up the ice to 0掳C and cooling down the water.
Latent Heat of Fusion
Latent heat of fusion is the amount of energy needed to change a substance from solid to liquid at a constant temperature. For ice, it involves absorbing heat without an obvious change in temperature, which is essential for this exercise.
When ice turns to water, it does so at 0掳C without the temperature increasing until the entire cube has melted. The latent heat of fusion for ice is:
  • Approximately 334,000 J/kg.
This value is necessary when calculating the total energy needed first to melt the ice before any further temperature rise for the resulting liquid water. Understanding this helps predict how the system's temperatures balance to reach thermal equilibrium after all these processes, particularly when deciding between using one or two ice cubes.

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Most popular questions from this chapter

What mass of butter, which has a usable energy content of \(6.0 \mathrm{Cal} / \mathrm{g}(=6000 \mathrm{cal} / \mathrm{g})\), would be equivalent to the change in gravitational potential energy of a \(73.0 \mathrm{~kg}\) man who ascends from sea level to the top of Mt. Everest, at elevation \(8.84 \mathrm{~km}\) ? Assume that the average \(g\) for the ascent is \(9.80 \mathrm{~m} / \mathrm{s}^{2}\).

(a) In 1964, the temperature in the Siberian village of Oymyakon reached \(-71^{\circ} \mathrm{C}\). What temperature is this on the Fahrenheit scale? (b) The highest officially recorded temperature in the continental United States was \(134^{\circ} \mathrm{F}\) in Death Valley, California. What is this temperature on the Celsius scale?

A small electric immersion heater is used to heat \(100 \mathrm{~g}\) of water for a cup of instant coffee. The heater is labeled "200 watts" (it converts electrical energy to thermal energy at this rate). Calculate the time required to bring all this water from \(23.0^{\circ} \mathrm{C}\) to \(100^{\circ} \mathrm{C}\),ignoring any heat losses.

In the extrusion of cold chocolate from a tube, work is done on the chocolate by the pressure applied by a ram forcing the chocolate through the tube. The work per unit mass of extruded chocolate is equal to \(p / \rho\), where \(p\) is the difference between the applied pressure and the pressure where the chocolate emerges from the tube, and \(\rho\) is the density of the chocolate. Rather than increasing the temperature of the chocolate, this work melts cocoa fats in the chocolate. These fats have a heat of fusion of \(150 \mathrm{~kJ} / \mathrm{kg}\). Assume that all of the work goes into that melting and that these fats make up \(30 \%\) of the chocolate's mass. What percentage of the fats melt during the extrusion if \(p=5.5\) MPa and \(\rho=1200 \mathrm{~kg} / \mathrm{m}^{3}\) ?

A \(1700 \mathrm{~kg}\) Buick moving at \(83 \mathrm{~km} / \mathrm{h}\) brakes to a stop, at uniform deceleration and without skidding, over a distance of \(93 \mathrm{~m}\). At what average rate is mechanical energy transferred to thermal energy in the brake system?
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