/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 7 In exercising, a weight lifter l... [FREE SOLUTION] | 91Ó°ÊÓ

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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. (b) Determine the minimum number of nutritional Calories of food (1 nutritional Calorie \(=4186 \mathrm{J}\) ) that must be consumed to replace the loss of internal energy.

Short Answer

Expert verified
Internal energy change is \(5.03 \times 10^5 \text{ J};\) consume ~120.2 Calories to replace it.

Step by step solution

01

Calculate the Energy Lost Due to Evaporation

To find the energy required for evaporating the water, use the latent heat of vaporization formula \( Q = mL \), where \( m \) is the mass of the water (\(0.150 \text{ kg}\)) and \( L \) is the latent heat (\(2.42 \times 10^6 \text{ J/kg}\)). Thus, \( Q = 0.150 \times 2.42 \times 10^6 = 3.63 \times 10^5 \text{ J} \).
02

Find the Total Change in Internal Energy

The total change in internal energy \( \Delta U \) is the sum of the energy lost due to evaporation and the work done lifting weights: \( \Delta U = Q + W \), where \( W = 1.40 \times 10^5 \text{ J} \). Thus, \( \Delta U = 3.63 \times 10^5 + 1.40 \times 10^5 = 5.03 \times 10^5 \text{ J} \).
03

Convert the Energy Loss to Nutritional Calories

To find the nutritional calories needed to replace the lost internal energy, use the conversion factor where \(1\) nutritional Calorie = \(4186 \text{ J}\). Thus, \( \frac{5.03 \times 10^5}{4186} \approx 120.2 \text{ Calories}\).
04

Conclusion

The change in internal energy of the weight lifter is \(5.03 \times 10^5 \text{ J}\) and they need to consume approximately 120.2 nutritional Calories to replace this energy loss.

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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 a fundamental concept in thermodynamics. It refers to the amount of energy required to turn a liquid into a vapor without changing its temperature. In this case, it is the energy needed for the water to evaporate from the weight lifter's body during exercise.
The formula for calculating this energy is straightforward:
  • \(Q = mL\)
  • where \(Q\) is the energy, \(m\) is the mass of the substance (in kg), and \(L\) is the latent heat of vaporization (in J/kg).
In the exercise, the weight lifter loses 0.150 kg of water. Using the given latent heat of vaporization, \(2.42 \times 10^6 \text{ J/kg}\), the energy required for evaporation is found to be \(3.63 \times 10^5 \text{ J}\).
Understanding this concept helps us appreciate the significant energy transfer that occurs even in a common process like sweating during exercise.
Internal Energy
Internal energy is an important concept in thermodynamics, as it represents the total energy contained within a system. It includes all forms of energy, such as kinetic and potential energy at the molecular level.
When the weight lifter exercises, they work to lift weights and lose energy through the evaporation of sweat. The change in internal energy (\( \Delta U \)) can be calculated by combining both these energy components:
  • The energy lost due to evaporation (\( 3.63 \times 10^5 \text{ J} \)
  • The work done lifting weights (\( 1.40 \times 10^5 \text{ J}\)
Using the formula:
  • \( \Delta U = Q + W \)
  • where \(Q\) is the energy lost through evaporation and \(W\) is the work done.
The total change in internal energy is found to be \(5.03 \times 10^5 \text{ J}\).
This illustrates how physical activities affect the body's energy balance.
Nutritional Calories
Nutritional calories are a crucial unit of measurement in understanding energy balance, especially when it comes to diet and exercise. In physics, energy is usually measured in joules, but the food industry uses nutritional calories to express energy values.
One nutritional Calorie is equivalent to 4186 joules, which helps to quantify the energy intake from food to replace the energy spent.
In this exercise, after calculating the total energy loss of the weight lifter as \(5.03 \times 10^5 \text{ J}\), it's essential to convert it into nutritional Calories to understand how much energy needs to be replenished through food.
  • The conversion is done using the formula:\( \text{Calories} = \frac{\text{Energy in J}}{4186} \)
Resultantly, the weight lifter needs to consume approximately 120.2 nutritional Calories to replace the lost energy.
This conversion is crucial for anyone looking to balance their energy expenditure with energy intake.

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Most popular questions from this chapter

Six grams of helium (molecular mass \(=4.0 \mathrm{u}\) ) expand isothermally at \(370 \mathrm{K}\) and does \(9600 \mathrm{J}\) of work. Assuming that helium is an ideal gas, determine the ratio of the final volume of the gas to the initial volume.

Beginning with a pressure of \(2.20 \times 10^{5}\) Pa and a volume of \(6.34 \times 10^{-3} \mathrm{m}^{3},\) an ideal monatomic gas \(\left(\gamma=\frac{5}{3}\right)\) undergoes an adiabatic expansion such that its final pressure is \(8.15 \times 10^{4} \mathrm{Pa}\). An alternative process leading to the same final state begins with an isochoric cooling to the final pressure, followed by an isobaric expansion to the final volume. How much more work does the gas do in the adiabatic process than in the alternative process?

The sublimation of zinc (mass per mole \(=0.0654 \mathrm{kg} / \mathrm{mol}\) ) takes place at a temperature of \(6.00 \times 10^{2} \mathrm{K},\) and the latent heat of sublimation is \(1.99 \times 10^{6} \mathrm{J} / \mathrm{kg}\). The pressure remains constant during the sublimation. Assume that the zinc vapor can be treated as a monatomic ideal gas and that the volume of solid zinc is negligible compared to the corresponding vapor. Concepts: (i) What is sublimation, and what is the latent heat of sublimation? (ii) When a solid phase changes to a gas phase, does the volume of the material increase or decrease, and by how much? (iii) As the material changes from a solid to a gas, does it do work on the environment, or does the environment do work on it? How much work is involved? (iv) In this problem we begin with heat \(Q\) and realize that it is used for two purposes: First, it makes the solid change into a gas, which entails a change \(\Delta U\) in the internal energy of the material, \(\Delta U=U_{\mathrm{gas}}-U_{\mathrm{solid}} .\) Second, it allows the expanding material to do work \(W\) on the environment. According to the conservation-of-energy principle, how is \(Q\) related to \(\Delta U\) and \(W ?\) (v) According to the first law of thermodynamics, how is \(Q\) related to \(\Delta U\) and \(W\) ? Calculations: What is the change in the internal energy of zinc when \(1.50 \mathrm{kg}\) of zinc sublimates?

A Carnot engine operates between temperatures of 650 and \(350 \mathrm{K}\). To improve the efficiency of the engine, it is decided either to raise the temperature of the hot reservoir by \(40 \mathrm{K}\) or to lower the temperature of the cold reservoir by \(40 \mathrm{K}\). Which change gives the greater improvement? Justify your answer by calculating the efficiency in each case.

One-half mole of a monatomic ideal gas absorbs \(1200 \mathrm{J}\) of heat while 2500 J of work is done by the gas. (a) What is the temperature change of the gas? (b) Is the change an increase or a decrease?

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