Question
(a) How much heat transfer occurs to the environment by an electrical power station that uses $1.25 \times 10^{14} \textrm{ J}$ of heat transfer into the engine with an efficiency of 42.0%? (b) What is the ratio of heat transfer to the environment to work output? (c) How much work is done?
1. $7.25 \times 10^{13} \textrm{ J}$
2. $1.38$
3. $5.25 \times 10^{13} \textrm{ J}$
Solution Video

# OpenStax College Physics Solution, Chapter 15, Problem 25 (Problems & Exercises) (3:43)

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Video Transcript
This is College Physics Answers with Shaun Dychko. A power station absorbs 1.24 times tenth to the 14 Joules of energy from the high temperature reservoir and it has an efficiency of 42 percent which is 0.420. In part A, we are asked to figure out how much energy is lost to the cold temperature reservoir so how much energy is just lost to the environment in other words. Well the efficiency is the work done by the power station divided by the energy that it absorbs, and we can rewrite the work as heat energy absorbed minus heat energy lost to the environment and then substitute that in place of W in our efficiency formula. So we can divide both terms and the numerator by this Qh in the denominator and that works out to one minus Qc over Qh and we are solving for Qc here and so we will add it to both sides and then subtract efficiency from both sides and we end up with Qc over Qh equals one minus efficiency and multiply both sides by Qh to solve for Qc. So Qc equals Qh times one minus efficiency. So, it’s 1.25 times tenth to the 14 Joules of energy absorbed multiply by one minus 0.42 efficiency. Giving us 7.25 times tenth to the 13 Joules of energy lost to the environment and then part B asks us to find the ratio of the energy lost to the environment to the amount of the energy produced usefully so the work done in other words. So, will substitute W as Qh minus Qc as we did up here. This is true for any cyclical process that has zero change in internal energy and so we also substitute for Qc and write it as Qh times one minus efficiency and now we have an expression containing only efficiency and as well as Qh but that’s gonna end up canceling as we'll see it’s a common factor in the top and bottom. So, we have one minus efficiency in the top divided by one minus one minus efficiency. So, this maybe, an intermediate step would look like this, it would be Qh times one minus efficiency divided by Qh times one minus one minus efficiency. So I’m showing the factoring out of Qh in the denominator there and there's one that appears here. So this bottom one minus one is zero and then you know minus minus efficiency makes positive efficiency in the denominator. So we’ve one minus efficiency divided by efficiency in other words. So that’s one minus 0.42 divided by 0.42 which is 1.38. So it’s 1.38 times as much energy lost to the environment as there is work done by the power station and now the work done is efficiency equals work done divided by Qh and we are gonna solve for work done by multiplying both sides by Qh and the work done is this 1.25 times tenth to the 14 Joules of energy absorbed times 0.42 which is 5.25 times tenth to the 13 Joules.