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Chapter 12: Problem 63

An unknown material has a normal melting/ freezing point of \(-25.0^{\circ} \mathrm{C},\) and the liquid phase has a specific heat capacity of \(160 \mathrm{~J} /\left(\mathrm{kg} \cdot \mathrm{C}^{\circ}\right)\). One-tenth of a kilogram of the solid at \(-25.0^{\circ} \mathrm{C}\) is put into a \(0.150-\mathrm{kg}\) aluminum calorimeter cup that contains 0.100 kg of glycerin. The temperature of the cup and the glycerin is initially \(27.0^{\circ} \mathrm{C}\). All the unknown material melts, and the final temperature at equilibrium is \(20.0^{\circ} \mathrm{C}\). The calorimeter neither loses energy to nor gains energy from the external environment. What is the latent heat of fusion of the unknown material?

Short Answer

Expert verified
The latent heat of fusion is approximately 19050 J/kg.

Step by step solution

01

Calculate Heat Lost by Glycerin and Calorimeter

First, identify the specific heat capacities needed: - Aluminum: \(c_{\text{Al}} = 900 \, \text{J/(kg路掳C)}\) - Glycerin: \(c_{\text{gly}} = 2400 \, \text{J/(kg路掳C)}\)Calculate the heat lost by glycerin (\(Q_{\text{gly}}\)):\[Q_{\text{gly}} = m_{\text{gly}} \cdot c_{\text{gly}} \cdot (T_{\text{initial}} - T_{\text{final}})\]\[Q_{\text{gly}} = 0.100 \, \text{kg} \cdot 2400 \, \text{J/(kg路掳C)} \cdot (27.0 掳C - 20.0 掳C)\]\[Q_{\text{gly}} = 1680 \, \text{J}\]Calculate the heat lost by the aluminum cup (\(Q_{\text{Al}}\)):\[Q_{\text{Al}} = m_{\text{Al}} \cdot c_{\text{Al}} \cdot (T_{\text{initial}} - T_{\text{final}})\]\[Q_{\text{Al}} = 0.150 \, \text{kg} \cdot 900 \, \text{J/(kg路掳C)} \cdot (27.0 掳C - 20.0 掳C)\]\[Q_{\text{Al}} = 945 \, \text{J}\]
02

Total Heat Lost by Cup and Glycerin

Add the heat lost by the glycerin and aluminum calorimeter cup:\[Q_{\text{total}} = Q_{\text{gly}} + Q_{\text{Al}}\]\[Q_{\text{total}} = 1680 \, \text{J} + 945 \, \text{J}\]\[Q_{\text{total}} = 2625 \, \text{J}\]
03

Heat Gained by Unknown Material

The heat gained by the unknown material equals the heat lost by the glycerin and aluminum, assuming no heat is lost to the environment:\[Q_{\text{unknown}} = Q_{\text{total}}\]\[Q_{\text{unknown}} = 2625 \, \text{J}\]
04

Calculate Latent Heat of Fusion

The heat gained by the material involves both the heat required to melt and change the temperature:The equation for the heat is:\[Q_{\text{unknown}} = m \cdot c_{\text{liq}} \cdot (T_{\text{final}} - T_{\text{melting}}) + m \cdot L_f\]Solve for the latent heat of fusion \(L_f\):\[2625 \, \text{J} = 0.1 \, \text{kg} \cdot 160 \, \text{J/(kg路掳C)} \cdot (20.0 掳C - (-25.0 掳C)) + 0.1 \, \text{kg} \cdot L_f\]Calculate the heat for the temperature change:\[0.1 \, \text{kg} \cdot 160 \, \text{J/(kg路掳C)} \cdot 45.0 掳C = 720 \, \text{J}\]Calculate \(L_f\):\[2625 \, \text{J} = 720 \, \text{J} + 0.1 \, \text{kg} \cdot L_f\]\[1905 \, \text{J} = 0.1 \, \text{kg} \cdot L_f\]\[L_f = \frac{1905 \, \text{J}}{0.1 \, \text{kg}} = 19050 \, \text{J/kg}\]
05

Conclusion

The latent heat of fusion of the unknown material is approximately 19050 J/kg.

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Key Concepts

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

calorimetry
Calorimetry is an experimental technique to measure the heat exchanged in physical and chemical processes. It is fundamental for understanding thermal interactions between different substances. In the provided problem, you use calorimetry to calculate the latent heat of fusion of an unknown material. The basic idea is to measure how much heat is exchanged as the material transitions from solid to liquid and reaches thermal equilibrium with its environment.

Here, you employ a calorimeter, which is an insulated device that minimizes heat exchange with the surroundings, allowing you to attribute any heat transfer to the substances inside the calorimeter. The calorimeter in the problem is made of aluminum and contains glycerin, each having specific heat capacities. These specific heat capacities help determine the total heat exchange involved.
  • First, calculate the heat lost by the glycerin and the aluminum calorimeter because they both cool down from 27.0 掳C to the equilibrium temperature of 20.0 掳C.
  • Second, determine the heat gained by the unknown material as it heats up and melts.
  • Using the data and the heat transfer calculations, you determine the latent heat of fusion for the material.
specific heat capacity
Specific heat capacity is the amount of heat required to change the temperature of a unit mass of a substance by one degree Celsius. This property is crucial when analyzing heat transfer processes such as those in the problem. Each material in the calorimeter setup, whether aluminum or glycerin, has a unique specific heat capacity. Knowing these values allows you to calculate how much energy is lost or gained as temperature changes occur.

In the original problem, two specific heat capacities are explicitly needed:
  • Aluminum, with a specific heat capacity of 900 J/(kg路掳C),
  • Glycerin, with a specific heat capacity of 2400 J/(kg路掳C).
These values tell you that glycerin requires more energy per kilogram per degree Celsius change than aluminum, making it a significant factor in heat exchange calculations.

When the unknown material melts inside the calorimeter, it also has a specific heat capacity in its liquid phase, given as 160 J/(kg路掳C). Use this to calculate how much heat the liquid gains until it reaches the final equilibrium temperature. This plays a role in quantifying the material's total heat gain throughout the process.
thermal equilibrium
Thermal equilibrium occurs when two or more substances in thermal contact with each other no longer exchange heat, achieving the same temperature. It's a vital concept because it allows you to make the assumption that total heat lost equals total heat gained within a closed system like a calorimeter.

Within the problem, thermal equilibrium is reached when the unknown material, the aluminum calorimeter, and the glycerin stabilize at a common temperature of 20.0 掳C. This is after all the heat transfers between them have occurred and no net heat flow takes place.
  • Initially, the glycerin and the aluminum cup are warmer at 27.0 掳C, while the unknown solid is much colder at -25.0 掳C.
  • As the process unfolds, heat flows from the warmer substances (glycerin and aluminum) to the colder one (the unknown material) until all entities balance at 20.0 掳C.
  • This steady state means that the system has reached thermal equilibrium, allowing the latent heat of fusion to be calculated based on the heat exchanges necessary to get to this point.

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Most popular questions from this chapter

Concept Questions (a) Two portions of the same liquid have the same mass, but different temperatures. They are mixed in a container that prevents the exchange of heat with the environment. Portion A has the higher initial temperature of \(T_{0 \mathrm{~A}},\) while portion \(\mathrm{B}\) has an initial temperature of \(T_{0 \mathrm{~B}}\). Relative to these two temperatures, where on the temperature scale will the final equilibrium temperature of the mixture be located? Explain. (b) Three portions of the same liquid are mixed in a container that prevents the exchange of heat with the environment. Portion A has a mass \(m\) and a temperature of \(94.0^{\circ} \mathrm{C}\), portion \(\mathrm{B}\) has a mass \(m\) and a temperature of \(78.0^{\circ} \mathrm{C},\) and portion \(\mathrm{C}\) has a mass \(2 \mathrm{~m}\) and a temperature of \(34.0{ }^{\circ} \mathrm{C}\). Considering your answer to Concept Question (a) and without doing a heat-lost-equals-heat-gained calculation, deduce the final temperature of the mixture at equilibrium. Justify your answer. Problem Three portions of the same liquid are mixed in a container that prevents the exchange of heat with the environment. Portion A has a mass \(m\) and a temperature of \(94.0^{\circ} \mathrm{C},\) portion \(\mathrm{B}\) has a mass \(m\) and a temperature of \(78.0{ }^{\circ} \mathrm{C},\) and portion \(\mathrm{C}\) has a mass \(m_{\mathrm{C}}\) and a temperature of \(34.0^{\circ} \mathrm{C}\). What must be the mass of portion \(\mathrm{C}\) such that the final temperature \(T_{\mathrm{f}}\) of the three-portion mixture is \(T_{\mathrm{f}}=50.0^{\circ} \mathrm{C} ?\) Express your answer in terms of \(m ;\) e.g., \(m_{C}=2.20 m\).

Suppose you are selling apple cider for two dollars a gallon when the temperature is \(4.0^{\circ} \mathrm{C} .\) The coefficient of volume expansion of the cider is \(280 \times 10^{6}\left(\mathrm{C}^{\circ}\right)^{-1} .\) If the expansion of the container is ignored, how much more money (in pennies) would you make per gallon by refilling the container on a day when the temperature is \(26^{\circ} \mathrm{C} ?\)

To help prevent frost damage, fruit growers sometimes protect their crop by spraying it with water when overnight temperatures are expected to go below the freezing mark. When the water turns to ice during the night, heat is released into the plants, thereby giving them a measure of protection against the falling temperature. Suppose a grower sprays \(7.2 \mathrm{~kg}\) of water at \(0^{\circ} \mathrm{C}\) onto a fruit tree. (a) How much heat is released by the water when it freezes? (b) How much would the temperature of a \(180-\mathrm{kg}\) tree rise if it absorbed the heat released in part (a)? Assume that the specific heat capacity of the tree is \(2.5 \times 10^{3} \mathrm{~J} /\left(\mathrm{kg} \cdot \mathrm{C}^{\circ}\right)\) and that no phase change occurs within the tree itself.

Occasionally, huge icebergs are found floating on the ocean's currents. Suppose one such iceberg is \(120 \mathrm{~km}\) long, \(35 \mathrm{~km}\) wide, and \(230 \mathrm{~m}\) thick. (a) How much heat would be required to melt this iceberg (assumed to be at \(0{ }^{\circ} \mathrm{C}\) ) into liquid water at \(0{ }^{\circ} \mathrm{C}\) ? The density of ice is \(917 \mathrm{~kg} / \mathrm{m}^{3}\). (b) The annual energy consumption by the United States in 1994 was \(9.3 \times 10^{19} \mathrm{~J}\). If this energy were delivered to the iceberg every year, how many years would it take before the ice melted?

Dermatologists often remove small precancerous skin lesions by freezing them quickly with liquid nitrogen, which has a temperature of \(77 \mathrm{~K}\). What is this temperature on the (a) Celsius and (b) Fahrenheit scales?
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