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Chapter 18: Problem 42

In one of your rigorous workout sessions, you lost \(150 \mathrm{~g}\) of water through evaporation. Assume that the amount of work done by your body was \(1.80 \cdot 10^{5} \mathrm{~J}\) and that the heat required to evaporate the water came from your body. a) Find the loss in internal energy of your body, assuming the latent heat of vaporization is \(2.42 \cdot 10^{6} \mathrm{~J} / \mathrm{kg}\). b) Determine the minimum number of food calories that must be consumed to replace the internal energy lost (1 food calorie \(=4186\) J).

Short Answer

Expert verified
Answer: 44 food calories.

Step by step solution

01

Calculate the energy required to evaporate the given amount of water

To calculate the energy required to evaporate the given amount of water, we use the formula: Energy = Mass x Latent heat of vaporization where Mass = \(150\,\text{g}\) (converted to kg) and Latent heat of vaporization = \(2.42 \cdot 10^6\, \mathrm{J} / \mathrm{kg}\). Converting the mass of water to kg: \(150\,\text{g} \times \frac{1\,\text{kg}}{1000\, \text{g}} = 0.15\, \text{kg}\). Energy = \(0.15\,\text{kg} \times 2.42 \cdot 10^6 \,\mathrm{J} / \mathrm{kg}\) Energy = \(3.63 \cdot 10^5\, \text{J}\)
02

Determine the loss in internal energy

According to the conservation of energy, the work done by the body is equal to the energy required to evaporate the water plus the change in the internal energy of your body, Work done = Energy + Loss in internal energy Rearranging the equation, we get: Loss in internal energy = Work done - Energy Loss in internal energy = \(1.80 \cdot 10^5\, \text{J} - 3.63 \cdot 10^5\, \text{J}\) Loss in internal energy = \(-1.83 \cdot 10^5\, \text{J}\) The negative sign indicates that the internal energy of your body has decreased. b) Determine the minimum number of food calories
03

Convert the Joules to food calories

To convert the lost energy from Joules to food calories, we use the conversion factor given, 1 food calorie = 4186 J Lost energy in food calories = \(\frac{-1.83 \cdot 10^5\, \text{J}}{4186\, \mathrm{J} / \mathrm{food\, calorie}}\) Lost energy in food calories = \(-43.77\, \text{food calories}\)
04

Find the minimum number of food calories needed to replace the lost internal energy

Since energy cannot be negative, we must consume at least 43.77 food calories to replace the lost internal energy. Minimum number of food calories = 44 (rounded up)

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Most popular questions from this chapter

The latent heat of vaporization of liquid nitrogen is about \(200 . \mathrm{kJ} / \mathrm{kg} .\) Suppose you have \(1.00 \mathrm{~kg}\) of liquid nitrogen boiling at \(77.0 \mathrm{~K}\). If you supply heat at a constant rate of \(10.0 \mathrm{~W}\) via an electric heater immersed in the liquid nitrogen, how long will it take to vaporize all of it? What is the time for \(1.00 \mathrm{~kg}\) of liquid helium, whose heat of vaporization is \(20.9 \mathrm{~kJ} / \mathrm{kg}\) ?

The radiation emitted by a blackbody at temperature \(T\) has a frequency distribution given by the Planck spectrum: $$ \epsilon_{T}(f)=\frac{2 \pi h}{c^{2}}\left(\frac{f^{3}}{e^{h f / k_{\mathrm{B}} T}-1}\right) $$ where \(\epsilon_{T}(f)\) is the energy density of the radiation per unit increment of frequency, \(v\) (for example, in watts per square meter per hertz), \(h=6.626 \cdot 10^{-34} \mathrm{~J} \mathrm{~s}\) is Planck's constant, \(k_{\mathrm{B}}=1.38 \cdot 10^{-23} \mathrm{~m}^{2} \mathrm{~kg} \mathrm{~s}^{-2} \mathrm{~K}^{-1}\) is the Boltzmann constant, and \(c\) is the speed of light in vacuum. (We'll derive this distribution in Chapter 36 as a consequence of the quantum hypothesis of light, but here it can reveal something about radiation. Remarkably, the most accurately and precisely measured example of this energy distribution in nature is the cosmic microwave background radiation.) This distribution goes to zero in the limits \(f \rightarrow 0\) and \(f \rightarrow \infty\) with a single peak in between those limits. As the temperature is increased, the energy density at each frequency value increases, and the peak shifts to a higher frequency value. a) Find the frequency corresponding to the peak of the Planck spectrum, as a function of temperature. b) Evaluate the peak frequency at temperature \(T=6.00 \cdot 10^{3} \mathrm{~K}\), approximately the temperature of the photosphere (surface) of the Sun. c) Evaluate the peak frequency at temperature \(T=2.735 \mathrm{~K}\), the temperature of the cosmic background microwave radiation. d) Evaluate the peak frequency at temperature \(T=300 . \mathrm{K}\), which is approximately the surface temperature of Earth.

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