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Chapter 15: Problem 7

In exercising, a weight lifter loses 0.150 \(\mathrm{kg}\) of water through evaporation, the heat required to evaporate the water coming from the weight lifter's body. The work done in lifting weights is \(1.40 \times 10^{5} \mathrm{J}\) (a) Assuming that the latent heat of vaporization of perspiration is \(2.42 \times 10^{6} \mathrm{J} / \mathrm{kg}\) find the change in the internal energy of the weight lifter. \((\mathbf{b})\) Determine the minimum number of nutritional Calories of food \((1 \text { nutritional Calorie }=4186 \mathrm{J})\) that must be consumed to replace the loss of internal energy.

Short Answer

Expert verified
(a) \( \Delta U = -5.03 \times 10^5 \, \text{J} \); (b) 120 nutritional Calories.

Step by step solution

01

Calculate the Heat Required for Evaporation

The heat required to evaporate 0.150 kg of water can be calculated using the latent heat of vaporization. The formula is \( Q = m \cdot L_v \), where \( Q \) is the heat, \( m \) is the mass of the water, and \( L_v \) is the latent heat of vaporization. Substituting the given values, we have: \( Q = 0.150\, \text{kg} \times 2.42 \times 10^6 \, \text{J/kg} = 3.63 \times 10^5 \, \text{J} \).
02

Use the First Law of Thermodynamics

According to the first law of thermodynamics, the change in internal energy \( \Delta U \) is given by \( \Delta U = Q - W \), where \( Q \) is the heat added, and \( W \) is the work done by the system. Here, \( Q = 3.63 \times 10^5 \, \text{J} \) (heat lost due to evaporation) and \( W = 1.40 \times 10^5 \, \text{J} \) (work done in lifting weights). So, \( \Delta U = -3.63 \times 10^5 \, \text{J} - 1.40 \times 10^5 \, \text{J} \).
03

Calculate Change in Internal Energy

Plugging the values in, we find the change in internal energy: \( \Delta U = -3.63 \times 10^5 \, \text{J} - 1.40 \times 10^5 \, \text{J} = -5.03 \times 10^5 \, \text{J} \).
04

Calculate Nutritional Calories Required

To find the nutritional Calories required to replace the lost internal energy, convert joules to nutritional Calories. Since \( 1 \) nutritional Calorie = \( 4186 \, \text{J} \), the equation is \( \frac{5.03 \times 10^5 \, \text{J}}{4186 \, \text{J/Calorie}} = 120.13 \; \text{Calories} \).

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Key Concepts

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

Latent Heat of Vaporization
The latent heat of vaporization is the amount of heat required to convert a unit mass of a liquid into vapor without a temperature change. For water, this is particularly important as it has a relatively high value of latent heat. In the context of our exercise, the weight lifter loses water due to sweating, which then evaporates. Using the equation for latent heat, you find that to evaporate 0.150 kg of water, a substantial amount of heat, namely 363,000 J, is required. This heat comes from the weight lifter's body, essentially cooling the body down during physical exertion.
First Law of Thermodynamics
The first law of thermodynamics is a principle that provides the concept of energy conservation in thermodynamic systems. It states that the change in the internal energy of a system (\( \Delta U \)) is equal to the heat added to the system (\( Q \)) plus the work done by the system (\( W \)). This can be expressed as: \[\Delta U = Q - W\].
In the weight lifter's situation, both evaporation and lifting weights contribute to the energy changes in the body:
  • Heat is removed by evaporation: \( -3.63 \times 10^5 \, \text{J} \)
  • Work is done in lifting weights: \( -1.40 \times 10^5 \, \text{J} \)
The net effect is a decrease in internal energy amounting to \( -5.03 \times 10^5 \, \text{J} \).
Energy Conversion
Energy conversion is crucial in understanding how different forms of energy transition from one type to another. In thermodynamics, you often deal with conversions between thermal energy and mechanical work. For the weight lifter, thermal energy in the form of body heat is converted to mechanical energy to perform the lifting work. Simultaneously, some of this thermal energy is used for the phase change of perspiration, which is the conversion of heat energy into latent heat. The efficiency of these conversions is part of understanding energy balance in physiological systems.
Caloric Intake Calculation
Understanding caloric intake is fundamental for maintaining energy balance in the body, especially for athletes. The exercise example involves calculating the replacement of lost energy through food intake, measured in nutritional Calories. A nutritional Calorie, a unit of energy consumption, equals 4,186 J.
To replace the 503,000 J lost due to exertion and evaporation, you convert joules into nutritional Calories:
  • Calculation: \( \frac{5.03 \times 10^5 \, \text{J}}{4186 \, \text{J/Calorie}} = 120.13 \text{ Calories} \)
This becomes the minimum caloric intake needed to compensate for the energy loss, ensuring the weight lifter maintains energy levels for further activities.

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Most popular questions from this chapter

Argon is a monatomic gas whose atomic mass is 39.9 u. The temperature of eight grams of argon is raised by 75 \(\mathrm{K}\) under conditions of constant pressure. Assuming that argon behaves as an ideal gas, how much heat is required?

A system undergoes a two-step process. In the first step, the internal energy of the system increases by 228 \(\mathrm{J}\) when 166 \(\mathrm{J}\) of work is done on the system. In the second step, the internal energy of the system increases by 115 \(\mathrm{J}\) when 177 \(\mathrm{J}\) of work is done on the system. For the overall process, find the heat. What type of process is the overall process? Explain.

Suppose that 31.4 \(\mathrm{J}\) of heat is added to an ideal gas. The gas expands at a constant pressure of \(1.40 \times 10^{4}\) Pa while changing its volume from \(3.00 \times 10^{-4}\) to \(8.00 \times 10^{-4} \mathrm{m}^{3}\) . The gas is not monatomic, so the relation \(C_{P}=\frac{5}{2} R\) does not apply. (a) Determine the change in the internal energy of the gas. (b) Calculate its molar specific heat capacity \(C_{p} .\)

When a .22 -caliber rifle is fired, the expanding gas from the burning gunpowder creates a pressure behind the bullet. This pressure causes the force that pushes the bullet through the barrel. The barrel has a length of 0.61 \(\mathrm{m}\) and an opening whose radius is \(2.8 \times 10^{-3} \mathrm{m} . \mathrm{A}\) bullet (mass \(=2.6 \times 10^{-3} \mathrm{kg}\) ) has a speed of 370 \(\mathrm{m} / \mathrm{s}\) after passing through this barrel. Ignore friction and determine the average pressure of the expanding gas.

A 52 -kg mountain climber, starting from rest, climbs a vertical distance of 730 \(\mathrm{m}\) . At the top, she is again at rest. In the process, her body generates \(4.1 \times 10^{6} \mathrm{J}\) of energy via metabolic processes. In fact, her body acts like a heat engine, the efficiency of which is given by Equation 15.11 as \(e=|W| / Q_{\mathrm{H}} |,\) where \(|W|\) is the magnitude of the work she does and \(\left|Q_{\mathrm{H}}\right|\) is the magnitude of the input heat. Find her efficiency as a heat engine.
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