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A person eats a container of strawberry yogurt. The Nutritional Facts label states that it contains 240 Calories ( 1 Calorie \(=4186\) 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 person needs to lose approximately 0.415 kg of perspiration.

Step by step solution

01

Understand the Problem

We need to determine the mass of perspiration needed to evaporate and burn 240 Calories (kcal) through vaporization at body temperature. This involves understanding that 1 Calorie is equivalent to 4186 joules of energy, and the latent heat of vaporization (the energy required to turn liquid into vapor) is given as \(2.42 \times 10^{6}\, \mathrm{J/kg}\).
02

Convert Calories to Joules

First, convert the energy content of the yogurt from Calories to Joules using the conversion factor: 1 Calorie = 4186 Joules. Multiply:\[240 \text{ Cal} \times 4186 \text{ J/Cal} = 1004640 \text{ J}\].
03

Use the Energy Equation

Use the equation to find mass \(m\) of perspiration: \[ E = m \times L \ \text{where } E \text{ is energy in joules and } L \text{ is the latent heat of vaporization.} \]Rearrange to solve for mass \(m\):\[ m = \frac{E}{L} = \frac{1004640}{2.42 \times 10^{6}} \, \text{kg} \]
04

Calculate the Mass of Perspiration

Substitute the values into the equation:\[ m = \frac{1004640}{2.42 \times 10^{6}} m \approx 0.415 \text{ kg} \ \text{Thus, the mass of perspiration required is approximately } 0.415 \text{ kg.}\]

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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 the human body, it's often measured in Calories. However, scientific calculations typically require energy to be in joules. Luckily, converting Calories to joules is straightforward. One dietary Calorie (with a capital 'C'), often just called "Calorie," is equivalent to 4186 joules. This conversion factor is crucial in any calculations involving energy release or usage in the body. To convert energy from Calories to joules, you simply multiply the number of Calories by 4186. For instance, if a serving of yogurt contains 240 Calories, the energy expressed in joules becomes: - 240 Calorie × 4186 J/Calorie = 1004640 joules. Using this conversion allows us to perform further calculations to understand how our bodies use energy.
Evaporation and Energy Loss
Our bodies lose energy through various processes. One of the most effective ways our body dissipates excess heat and energy is by evaporating sweat. This process not only cools us down but is also a significant mechanism for energy loss.The body utilizes the latent heat of vaporization for this process. This is the amount of energy required to convert a liquid (sweat, in this case) to a gas (vapor) without changing temperature. For water, which makes up most of our sweat, the latent heat of vaporization at body temperature is approximately \(2.42 \times 10^{6} \mathrm{J} / \mathrm{kg}\). This high energy requirement is why sweating is such an effective way to lose heat.Therefore, when performing activities that make us sweat, our bodies are actively using up this energy to convert sweat into vapor, which leaves the skin surface cooler.
Mass of Perspiration
Determining the mass of perspiration needed to rid the body of a given amount of energy involves a straightforward calculation. It's essential when you consider scenarios like burning the Calories consumed through food via exercise or other activities.Let's revisit our example: how much sweat must evaporate to dissipate 1004640 joules, which is the energy from 240 dietary Calories? Using the formula:- \( E = m \times L \)Where \( E \) is the energy in joules, \( m \) is the mass of perspiration, and \( L \) is the latent heat of vaporization.We solve for \( m \):- \( m = \frac{E}{L} = \frac{1004640}{2.42 \times 10^{6}} \approx 0.415 \text{ kg} \).This tells us that approximately 0.415 kg of sweat must evaporate to release the energy contained in that yogurt serving. Engaging in activities that cause you to sweat can be a practical application of this calculation! By understanding and utilizing energy conversion and loss, we gain a better grip on energy balance in relation to physical activities.

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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 \(0.100 \mathrm{kg}\) 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?

(a) Objects A and B have the same mass of \(3.0 \mathrm{kg}\). They melt when \(3.0 \times 10^{4} \mathrm{J}\) of heat is added to object \(\mathrm{A}\) and when \(9.0 \times 10^{4} \mathrm{J}\) is added to object B. Determine the latent heat of fusion for the substance from which each object is made. (b) Find the heat required to melt object A when its mass is \(6.0 \mathrm{kg}\).

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.

You and your team are given the task of constructing a crude thermometer that covers a temperature range from \(0^{\circ} \mathrm{C}\) to \(60^{\circ} \mathrm{C} .\) You have at your disposal an aluminum rod of length \(L_{0}=2.00 \mathrm{m}\) (at \(\left.T=0^{\circ} \mathrm{C}\right)\) and diameter \(D=0.50 \mathrm{cm},\) and a broken clock that is missing its hour hand, and its longer minute hand is hanging loose and pointing in the six o"clock direction. The long hand of the clock is 8.20 inches long and pivots loosely at the clock's center. You get the idea to mount the rod horizontally so that one end butts against a wall and the other end pushes against the dangling minute hand of the clock. A temperature-induced change in the length of the rod will then be reflected in a change in the angle that the minute hand makes with the vertical (i.e., relative to the six o'clock position). (a) If you want the full temperature range to span the angle between the 6 and 7 markings (uniformly spaced) on the clock, how far from the central pivot should the end of the rod make contact with the minute hand? (b) The current temperature in the room is \(65.0^{\circ} \mathrm{F} .\) At what angle relative to the vertical (the six o'clock position) should the minute hand point at this temperature if it is to point directly at 6 when \(T=0^{\circ} \mathrm{C} ?\) (c) What would be the angular range of your "clock-thermometer" if the rod were made of steel, rather than aluminum? Assume that it is placed at the same position on the minute hand as determined in (a).

A 42 -kg block of ice at \(0^{\circ} \mathrm{C}\) is sliding on a horizontal surface. The initial speed of the ice is \(7.3 \mathrm{m} / \mathrm{s}\) and the final speed is \(3.5 \mathrm{m} / \mathrm{s} .\) Assume that the part of the block that melts has a very small mass and that all the heat generated by kinetic friction goes into the block of ice. Determine the mass of ice that melts into water at \(0^{\circ} \mathrm{C}\).

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