/*! 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 130 A man ate 0.50 pound of cheese (... [FREE SOLUTION] | 91Ó°ÊÓ

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A man ate 0.50 pound of cheese (an energy intake of \(4 \times 10^{3} \mathrm{~kJ}\) ). Suppose that none of the energy was stored in his body. What mass (in grams) of water would he need to perspire in order to maintain his original temperature? (It takes \(44.0 \mathrm{~kJ}\) to vaporize 1 mole of water.)

Short Answer

Expert verified
1,636 grams of water need to be vaporized.

Step by step solution

01

Understanding the Problem

To solve this problem, we need to find out how much water must be vaporized to get rid of the 4000 kJ of energy consumed from the cheese. This requires calculating the number of moles of water that would need to be vaporized using the heat of vaporization constant.
02

Calculate Moles of Water Needed

First, determine the number of moles of water needed to dissipate 4000 kJ of energy. Since it takes 44.0 kJ to vaporize 1 mole of water, we use the equation: \[ \text{moles of water} = \frac{4000 \text{ kJ}}{44.0 \text{ kJ/mole}} \] Calculate this to find the number of moles.
03

Perform the Calculation

Substitute 4000 kJ and 44.0 kJ/mole into the equation and calculate: \[\text{moles of water} = \frac{4000}{44.0} \approx 90.91 \text{ moles}\] This tells us how many moles of water need to evaporate to release the 4000 kJ of energy.
04

Convert Moles to Mass

Next, we convert moles of water to grams. The molecular weight of water (Hâ‚‚O) is 18.0 g/mole. So, the mass in grams is calculated by using the equation: \[\text{mass of water} = 90.91 \text{ moles} \times 18.0 \text{ g/mole}\] Calculate this to find the mass in grams.
05

Final Calculation

Complete the calculation:\[\text{mass of water} = 90.91 \times 18.0 = 1636.38 \text{ grams}\] This gives the required mass of water that needs to be vaporized to maintain body temperature.

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Key Concepts

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

Energy Dissipation
When we talk about energy dissipation, especially in the context of human body processes, it's all about the energy balance. Our body takes in energy through food, like the 0.50 pound of cheese in the exercise, which gives an energy intake of 4000 kJ. This energy can be utilized for various bodily functions or dissipated in other forms. If it's not stored, the excess energy needs to be expelled to maintain balance, keeping constant parameters like body temperature.

In this scenario, the solution involves dissipating energy through perspiration. This is a natural process our bodies use to regulate temperature. Excess body heat converts our perspiration (sweat) into vapor, using the energy absorbed from the body and then releasing it as vapor into the air. The key takeaway here is that the dissipation requires energy, which is captured by the heat of vaporization of water.
Thermodynamics
Thermodynamics is the study of energy transfer and changes in matter caused by this energy. In this context, the energy from the cheese you consume must follow the laws of thermodynamics. This means understanding how the energy intake (4000 kJ, in this case) affects body processes. The law pertinent here is the First Law of Thermodynamics, which tells us that energy cannot be created or destroyed; it can only be changed from one form to another.

The problem revolves around how the body expends this energy while maintaining thermal equilibrium. Since no energy is stored as per the prompt, and assuming the body maintains its temperature, the excess energy is entirely used to convert liquid sweat into vapor. The heat of vaporization is crucial here, where it defines how much energy is needed to convert liquid water into vapor. This principle links back to why calculating the mass of water dissipated is vital to understanding thermodynamics in our daily life.
Mole Calculation
In chemistry, the mole is a fundamental unit used for measuring amount of substance. Calculating moles is essential for understanding reactions and processes involving substances, like how much water needs to evaporate here to dissipate energy.

To determine the number of moles of water required, we need two main pieces of information:
  • The amount of energy to be dissipated, which is 4000 kJ in our exercise.
  • The heat of vaporization for water, which is 44.0 kJ per mole.
The calculation involves dividing the total energy by the energy needed per mole to vaporize water:\[\text{moles of water} = \frac{4000 \text{ kJ}}{44.0 \text{ kJ/mole}} = 90.91 \text{ moles}\]Once we know the moles of water, converting these moles into grams helps understand the actual mass of water. Since the molecular weight of water is 18.0 g/mole, we multiply by the number of moles:\[\text{mass of water} = 90.91 \times 18.0 = 1636.38 \text{ grams}\]Understanding these mole calculations helps you grasp how chemical principles apply in physiological processes, like energy dissipation through sweating.

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Most popular questions from this chapter

Consider the reaction $$\begin{aligned}2 \mathrm{H}_{2} \mathrm{O}(g) \longrightarrow & 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \\ \Delta H=&+483.6 \mathrm{~kJ} / \mathrm{mol}\end{aligned}$$ at a certain temperature. If the increase in volume is 32.7 \(\mathrm{L}\) against an external pressure of \(1.00 \mathrm{~atm},\) calculate \(\Delta U\) for this reaction. \((1 \mathrm{~L} \cdot \mathrm{atm}=101.3 \mathrm{~J})\)

Which of the following standard enthalpy of formation values is not zero at \(25^{\circ} \mathrm{C}: \mathrm{Na}(\) monoclinic \(), \mathrm{Ne}(g)\) \(\mathrm{CH}_{4}(g), \mathrm{S}_{8}(\) monoclinic \(), \mathrm{Hg}(l), \mathrm{H}(g) ?\)

Lime is a term that includes calcium oxide \((\mathrm{CaO},\) also called quicklime) and calcium hydroxide \(\left[\mathrm{Ca}(\mathrm{OH})_{2}\right.\) also called slaked lime]. It is used in the steel industry to remove acidic impurities, in air-pollution control to remove acidic oxides such as \(\mathrm{SO}_{2}\), and in water treatment. Quicklime is made industrially by heating limestone \(\left(\mathrm{CaCO}_{3}\right)\) above \(2000^{\circ} \mathrm{C}:\) \(\begin{aligned} \mathrm{CaCO}_{3}(s) \longrightarrow \mathrm{CaO}(s)+\mathrm{CO}_{2}(g) & \Delta H^{\circ}=177.8 \mathrm{~kJ} / \mathrm{mol} \end{aligned}\) Slaked lime is produced by treating quicklime with water: \(\mathrm{CaO}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(s)_{\Delta H^{\circ}}=-65.2 \mathrm{~kJ} / \mathrm{mol}\) The exothermic reaction of quicklime with water and the rather small specific heats of both quicklime \(\left[0.946 \mathrm{~J} /\left(\mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\right]\) and slaked lime \(\left[1.20 \mathrm{~J} /\left(\mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\right]\) make it hazardous to store and transport lime in vessels made of wood. Wooden sailing ships carrying lime would occasionally catch fire when water leaked into the hold. (a) If a 500.0 -g sample of water reacts with an equimolar amount of \(\mathrm{CaO}\) (both at an initial temperature of \(\left.25^{\circ} \mathrm{C}\right)\), what is the final temperature of the product, \(\mathrm{Ca}(\mathrm{OH})_{2} ?\) Assume that the product absorbs all the heat released in the reaction. (b) Given that the standard enthalpies of formation of \(\mathrm{CaO}\) and \(\mathrm{H}_{2} \mathrm{O}\) are -635.6 and \(-285.8 \mathrm{~kJ} / \mathrm{mol}\), respectively, calculate the standard enthalpy of formation of \(\mathrm{Ca}(\mathrm{OH})_{2}\).

A quantity of \(50.0 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) is mixed with \(50.0 \mathrm{~mL}\) of \(0.400 \mathrm{M} \mathrm{HNO}_{3}\) in a constant-pressure calorimeter having a heat capacity of \(496 \mathrm{~J} /{ }^{\circ} \mathrm{C}\). The initial temperature of both solutions is the same at \(22.4^{\circ} \mathrm{C}\). What is the final temperature of the mixed solution? Assume that the specific heat of the solutions is the same as that of water and the molar heat of neutralization is \(-56.2 \mathrm{~kJ} / \mathrm{mol}\).

A 44.0-g sample of an unknown metal at \(99.0^{\circ} \mathrm{C}\) was placed in a constant-pressure calorimeter containing \(80.0 \mathrm{~g}\) of water at \(24.0^{\circ} \mathrm{C}\). The final temperature of the system was found to be \(28.4^{\circ} \mathrm{C}\). Calculate the specific heat of the metal. (The heat capacity of the calorimeter is \(12.4 \mathrm{~J} /{ }^{\circ} \mathrm{C} .\)

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