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A person eats a container of strawberry yogurt. The Nutritional Facts label states that it contains 240 Calories ( 1 Calorie \(=4186 \mathrm{~J}\) ). What mass of perspiration would one have to lose to get rid of this energy? At body temperature, the latent heat of vaporization of water is \(2.42 \times 10^{6} \mathrm{~J} / \mathrm{kg} .\)

Short Answer

Expert verified
The mass of perspiration needed is approximately 0.415 kg.

Step by step solution

01

Convert Calories to Joules

First, we need to convert the energy from Calories to Joules. Since 1 Calorie is equivalent to 4186 Joules, we multiply the given 240 Calories by 4186 Joules per Calorie. \[ 240 ext{ Calories} \times 4186 \frac{\text{J}}{\text{Calorie}} = 1,004,640 \text{ J} \]
02

Use the Latent Heat Formula

Next, use the formula for latent heat to find the mass of the perspiration. The formula is: \[ Q = mL \] where \( Q \) is the energy lost, \( m \) is the mass of the sweat, and \( L \) is the latent heat of vaporization. We need to solve for \( m \) by rearranging the formula: \[ m = \frac{Q}{L} \]
03

Substitute the Values

Now substitute the known values into the equation: \[ m = \frac{1,004,640 \text{ J}}{2.42 \times 10^{6} \, \frac{\text{J}}{\text{kg}}} \]
04

Calculate the Mass

Perform the calculation to find \( m \):\[ m \approx \frac{1,004,640}{2,420,000} \approx 0.415 \text{ kg} \]

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Key Concepts

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

Calorie to Joule Conversion
When dealing with energy in food and nutrition labels, it's often listed in Calories. However, in scientific calculations, especially when converting energy into work or heat, we usually use Joules.
To switch between these two units, we use the conversion factor where 1 Calorie is equivalent to 4186 Joules. So, if you ever need to convert Calories into Joules, you simply multiply the number of Calories by 4186.
For example, in our exercise, we have 240 Calories from a yogurt. We do the math:
  • 240 Calories
  • Multiply by 4186 Joules per Calorie
That gives us a total energy content of 1,004,640 Joules. This conversion is crucial as it expresses the energy in a unit that allows us to further apply physics formulas.
Latent Heat of Vaporization
The latent heat of vaporization refers to the amount of energy needed to turn a substance from liquid into vapor at constant temperature. For water, this is particularly of interest as it is a common substance that undergoes phase changes.
At body temperature, water has a latent heat of vaporization of 2.42 x 10^6 J/kg. This means that it requires 2.42 million Joules of energy to convert one kilogram of water at body temperature into vapor.
This concept is fundamental because it helps us understand the energy required for processes like sweating, which is the body’s way of cooling itself. The heat energy absorbed by the body to vaporize the water (sweat) helps maintain a stable internal temperature.
Mass of Perspiration
When our body generates heat, it uses sweat to keep cool. This involves using energy to evaporate the sweat from the skin. To find out how much sweat or perspiration is required to dissipate a specific amount of energy, we use the formula relating energy, mass, and latent heat:
  • \( Q = mL \)
  • Where \( Q \) is the energy, \( m \) is the mass, and \( L \) is the latent heat of vaporization.
By rearranging this formula to solve for mass, \( m = \frac{Q}{L} \), we can substitute the known values. In the yogurt example, we found earlier that the energy was 1,004,640 Joules, and the latent heat of vaporization is 2.42 x 10^6 J/kg.
Thus, substituting these values in, we calculate the mass of perspiration to be approximately 0.415 kg. This mass tells us the amount of sweat needed to release the energy gained from the yogurt through evaporation.

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Most popular questions from this chapter

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?

An electric hot water heater takes in cold water at \(13.0^{\circ} \mathrm{C}\) and delivers hot water. The hot water has a constant temperature of \(45.0^{\circ} \mathrm{C}\) when the "hot" faucet is left open all the time, and the volume flow rate is \(5.0 \times 10^{-6} \mathrm{~m}^{3 / \mathrm{s}}\). What is the minimum power rating of the hot water heater?

A temperature of absolute zero occurs at \(-273.15^{\circ} \mathrm{C}\). What is this temperature on the Fahrenheit scale?

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?

A thermos contains \(150 \mathrm{~cm}^{3}\) of coffee at \(85^{\circ} \mathrm{C}\). To cool the coffee, you drop two \(11-\mathrm{g}\) ice cubes into the thermos. The ice cubes are initially at \(0^{\circ} \mathrm{C}\) and melt completely. What is the final temperature of the coffee? Treat the coffee as if it were water.

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