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A Carnot engine works between 880 K and 286 K and gives out 12 kW of mechanical power…

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A Carnot engine works between $880\ \text{K}$ and $286\ \text{K}$ and gives out $12\ \text{kW}$ of mechanical power. Find the heat it takes in and gives out each second. By how much does the entropy of each reservoir change each second?

Need a reminder? The key ideaShow

A Carnot engine is the best possible between two temperatures: $\eta = 1 - \dfrac{T_c}{T_h}$ and $\dfrac{Q_c}{Q_h} = \dfrac{T_c}{T_h}$ . The best refrigerator has $\text{COP} = \dfrac{T_c}{T_h - T_c}$ . Temperatures in kelvin.

Heat $Q$ taken in slowly at a steady temperature $T$ raises the entropy by $\Delta S = \dfrac{Q}{T}$ . Warming from $T_1$ to $T_2$ : $\Delta S = mc\ln\dfrac{T_2}{T_1}$ . A gas at a steady temperature: $\Delta S = nR\ln\dfrac{V_2}{V_1}$ . Entropy depends only on the state, and the total for an isolated system never falls.

The worked solution

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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

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