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Chapter 12: Problem 72

Equal masses of two different liquids have the same temperature of \(25.0^{\circ} \mathrm{C} .\) Liquid \(\mathrm{A}\) has a freezing point of \(-68.0^{\circ} \mathrm{C}\) and a specific heat capacity of \(1850 \mathrm{J} /\left(\mathrm{kg} \cdot \mathrm{C}^{\circ}\right) .\) Liquid \(\mathrm{B}\) has a freezing point of \(-96.0^{\circ} \mathrm{C}\) and a specific heat capacity of \(2670 \mathrm{J} /\left(\mathrm{kg} \cdot \mathrm{C}^{\circ}\right) .\) The same amount of heat must be removed from each liquid in order to freeze it into a solid at its respective freezing point. Determine the difference \(L_{\mathrm{f}, \mathrm{A}}-L_{\mathrm{f}, \mathrm{B}}\) between the latent heats of fusion for these liquids.

Short Answer

Expert verified
The latent heat difference is 151020 J/kg.

Step by step solution

01

Calculate Temperature Change for Each Liquid

For Liquid A, the temperature change is \[\Delta T_A = 25.0 - (-68.0) = 93.0^{\circ} \text{C}.\]For Liquid B, the temperature change is \[\Delta T_B = 25.0 - (-96.0) = 121.0^{\circ} \text{C}.\]
02

Calculate Heat Required for Cooling Each Liquid

The heat required to cool each liquid to its freezing point is given by \[Q = m \cdot c \cdot \Delta T.\]Assuming equal masses (\(m\)) for both:\[Q_A = m \cdot 1850 \cdot 93.0\]\[Q_B = m \cdot 2670 \cdot 121.0\]
03

Set Total Heat Removed Equal for Both Liquids

Since the same amount of heat is removed from each liquid:\[m \cdot 1850 \cdot 93.0 + m \cdot L_{f,A} = m \cdot 2670 \cdot 121.0 + m \cdot L_{f,B}.\]
04

Solve for Latent Heat Difference

Simplifying the equation, since masses are equal, they can be canceled out:\[1850 \cdot 93.0 + L_{f,A} = 2670 \cdot 121.0 + L_{f,B}\]\[L_{f,A} - L_{f,B} = 2670 \cdot 121.0 - 1850 \cdot 93.0\]\[L_{f,A} - L_{f,B} = 323070 - 172050\]\[L_{f,A} - L_{f,B} = 151020 \, \text{J/kg}.\]
05

Conclude with the Latent Heat Difference

The difference in latent heats of fusion between the two liquids is \(151020 \, \text{J/kg}\).

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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 tells us how much energy is needed to change the temperature of a material. In other words, it measures how much heat we must add or remove to increase or decrease the temperature of a substance by 1掳C for a mass of 1 kg.
In our exercise, we compare two liquids: Liquid A and Liquid B. Liquid A has a specific heat capacity of 1850 J/(kg路掳C), while Liquid B's specific heat capacity is higher at 2670 J/(kg路掳C).
  • A higher specific heat capacity means the liquid can store more heat energy before its temperature increases.
  • This tells us that Liquid B requires more energy than Liquid A to achieve the same temperature change.

Understanding specific heat capacity is essential for solving our exercise, where equal heat removal leads to cooling each liquid to its freezing point.
Temperature Change
Temperature change (\(\Delta T\)) is the difference between the initial and final temperatures. It helps us know how much a substance's temperature needs to drop or increase.
For Liquid A and Liquid B, the temperature change is calculated from their initial common temperature of \(25.0^{\circ}\)C to their respective freezing points (-68.0掳C for Liquid A, -96.0掳C for Liquid B).
  • Liquid A's temperature change: \(\Delta T_A = 25.0 - (-68.0) = 93.0^{\circ}\)C.
  • Liquid B's temperature change: \(\Delta T_B = 25.0 - (-96.0) = 121.0^{\circ}\)C.

This calculation reveals that Liquid B requires a greater temperature reduction than Liquid A to reach its freezing point. Understanding temperature change is crucial for determining how much energy, through heat, needs to be exchanged.
Freezing Point Depression
Freezing point depression refers to the lowering of a liquid's initial freezing point. In our context, it helps us understand how much we need to cool the liquids down to become solid.
Both liquids start at \(25.0^{\circ} \text{C}\). Liquid A freezes at \(-68.0^{\circ} \text{C}\), and Liquid B at \(-96.0^{\circ} \text{C}\). The original temperatures and their respective freezing points influence the amount of heat needed to drive these liquids to solidify.
  • Liquid A's freezing point depression is from \(25.0^{\circ} \text{C}\) to \(-68.0^{\circ} \text{C}\).
  • Liquid B's freezing point depression is from \(25.0^{\circ} \text{C}\) to \(-96.0^{\circ} \text{C}\).

Knowing the freezing point depression helps us in understanding why more or less energy (in the form of heat) must be removed for different substances. Combined with specific heat capacity and temperature change, it provides a fuller picture of the energy dynamics involved in processes like solidification.

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Most popular questions from this chapter

Multiple-Concept Example 4 reviews the concepts that are in volved in this problem. A ruler is accurate when the temperature is \(25^{\circ} \mathrm{C}\). When the temperature drops to \(-14^{\circ} \mathrm{C},\) the ruler shrinks and no longer measures distances accurately. However, the ruler can be made to read correctly if a force of magnitude \(1.2 \times 10^{3} \mathrm{N}\) is applied to each end so as to stretch it back to its original length. The ruler has a cross-sectional area of \(1.6 \times 10^{-5} \mathrm{m}^{2},\) and it is made from a material whose coefficient of linear expansion is \(2.5 \times 10^{-5}\left(\mathrm{C}^{\circ}\right)^{-1} .\) What is Young's modulus for the material from which the ruler is made?

One ounce of a well-known breakfast cereal contains 110 Calories (1 food Calorie \(=4186 \mathrm{J}\) ). If \(2.0 \%\) of this energy could be converted by a weight lifter's body into work done in lifting a barbell, what is the heaviest barbell that could be lifted a distance of \(2.1 \mathrm{m} ?\)

Ideally, when a thermometer is used to measure the temperature of an object, the temperature of the object itself should not change. However, if a significant amount of heat flows from the object to the thermometer, the temperature will change. A thermometer has a mass of\(31.0 \mathrm{g},\) a specific heat capacity of \(c=815 \mathrm{J} /\left(\mathrm{kg} \cdot \mathrm{C}^{\circ}\right),\) and a temperature of \(12.0^{\circ} \mathrm{C} .\) It is immersed in \(119 \mathrm{g}\) of water, and the final temperature of the water and thermometer is \(41.5^{\circ} \mathrm{C} .\) What was the temperature of the water before the insertion of the thermometer?

To help prevent frost damage, fruit growers sometimes protect their crop by spraying it with water when overnight temperatures are expected to go below freezing. When the water turns to ice during the night, heat is released into the plants, thereby giving a measure of protection against the cold. 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.

If a nonhuman civilization were to develop on Saturn's largest moon, Titan, its scientists might well devise a temperature scale based on the properties of methane, which is much more abundant on the surface than water is. Methane freezes at \(-182.6^{\circ} \mathrm{C}\) on Titan, and boils at \(-155.2^{\circ} \mathrm{C} .\) Taking the boiling point of methane as \(100.0^{\circ} \mathrm{M}\) (degrees Methane) and its freezing point as \(0^{\circ} \mathrm{M},\) what temperature on the Methane scale corresponds to the absolute zero point of the Kelvin scale?
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