A certain power plant puts out 580 MW of electric power. Estimate the heat discharged per second, assuming that the plant has an efficiency of 32%.
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This is Giancoli Answers with Mr. Dychko. Efficiency is the work done divided by the heat absorbed from the high temperature reservoir. We need to rewrite this in terms of QL the amount of heat discharged to the lower temperature reservoir and we know that work is QH minus QL . And we can bring this to the left hand side. In which case QH is work plus QL and so let me substitute that for QH . And we have efficiency is W over W plus QL and multiply both sides by that denominator W plus QL let me get this line here. Efficiency times that equals W and distribute the efficiency into the brackets there and we get efficiency tends to work plus efficiency times heat given after the low temperature reservoir equals the work down and then put this term on the other side and you get e times QL is W minus eW and then divide both sides by e and then also factor of the W because it looks nicer and you have work times one minus efficiency divided by efficiency is the amount of heat discharged to the low temperature reservoir. Now if you plug in the Watts here which is power instead of joules the only thing that will happen is that your answer will also be in in Watts or megawatts as I've written here. I mean strictly speaking I guess we should say that where we're dividing both sides by time here and one second but that's fine. So we have a 580 times ten to the six watts of work done which is the amount of electrical power that's supplied versus some other factors to do some work or something times one minus the efficiency of 0.32 divided by 0.32 gives this huge number and we'll write it move it’s decimal point over three times to make this ten to the six in which case we can call this megawatts. So this is 1200 megawatts.
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