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Chapter 5: Problem 18

A watt is a measure of power (the rate of energy change) equal to \(1 \mathrm{~J} / \mathrm{s}\). (a) Calculate the number of joules in a kilowatt- hour. (b) An adult person radiates heat to the surroundings at about the same rate as a 100 -watt electric incandescent lightbulb. What is the total amount of energy in kcal radiated to the surroundings by an adult in 24 hours?

Short Answer

Expert verified
(a) There are 3.6 million joules in a kilowatt-hour. (b) An adult person radiates approximately 2064.81 kilocalories of energy to the surroundings in a 24-hour period.

Step by step solution

01

Convert kilowatt-hours to watt-seconds

To convert kilowatt-hours to watt-seconds, we first need to convert kilowatts to watts. There are 1000 watts in 1 kilowatt, and there are 3600 seconds in 1 hour. So, 1 kilowatt-hour is equal to 1000 watts x 3600 seconds. 1 kilowatt-hour = 1000 watts × 3600 seconds
02

Convert watt-seconds to joules

Given that 1 watt is equal to 1 joule/second, we can use this relationship to convert watt-seconds to joules. Since 1 watt = 1 joule/second, 1 watt-second = 1 joule. 1 kilowatt-hour = 1000 watts × 3600 seconds × 1 joule/watt-second
03

Calculate the number of joules

We can now calculate the number of joules in a kilowatt-hour. 1 kilowatt-hour = 1000 × 3600 joules 1 kilowatt-hour = 3,600,000 joules So, there are 3.6 million joules in a kilowatt-hour. #b# Part (b): Calculate Energy in kcal radiated by an Adult in 24 Hours
04

Calculate energy in joules

The problem states that an adult person radiates heat at the rate of a 100-watt lightbulb. We are asked to calculate the total energy radiated in a 24-hour period. First, let's find out how many joules of energy are radiated in 24 hours: 100 watts × 24 hours × 3600 seconds/hour = 8,640,000 joules
05

Convert joules to kilocalories

Now, we need to convert joules to kilocalories. 1 kilocalorie is equal to 4184 joules. Therefore, we can use this conversion factor to calculate the energy in kilocalories: 8,640,000 joules × \( \frac{1 kcal}{4184 J} \)
06

Calculate the total energy in kcal

Finally, we can calculate the total energy in kilocalories: \( \frac{8,640,000}{4184} \) kcal ≈ 2064.81 kcal So, an adult person radiates approximately 2064.81 kilocalories of energy to the surroundings in a 24-hour period.

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Most popular questions from this chapter

Burning methane in oxygen can produce three different carbon-containing products: soot (very fine particles of graphite), \(\mathrm{CO}(g),\) and \(\mathrm{CO}_{2}(g)\) (a) Write three balanced equations for the reaction of methane gas with oxygen to produce these three products. In each case assume that \(\mathrm{H}_{2} \mathrm{O}(l)\) is the only other product. (b) Determine the standard enthalpies for the reactions in part (a). (c) Why, when the oxygen supply is adequate, is \(\mathrm{CO}_{2}(g)\) the predominant carbon-containing product of the combustion of methane?

Calcium carbide \(\left(\mathrm{CaC}_{2}\right)\) reacts with water to form acetylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\) and \(\mathrm{Ca}(\mathrm{OH})_{2}\). From the following enthalpy of reaction data and data in Appendix C, calculate \(\Delta H_{f}^{\circ}\) for \(\mathrm{CaC}_{2}(s):\) $$ \begin{aligned} \mathrm{CaC}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(s)+\mathrm{C}_{2} \mathrm{H}_{2}(g) \\ \Delta H^{\circ}=-127.2 \mathrm{~kJ} \end{aligned} $$

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