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 and and gives out 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?
A Carnot engine is the best possible between two temperatures: and . The best refrigerator has . Temperatures in kelvin.
Heat taken in slowly at a steady temperature raises the entropy by . Warming from to : . A gas at a steady temperature: . Entropy depends only on the state, and the total for an isolated system never falls.
Have you tried it?
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.
Step 1. Efficiency. , or .
Step 2. Heat in. .
Step 3. Heat out. .
Step 4. Entropy. The cold reservoir gains , and the hot one loses .
Answer: in, out; each reservoir changes by , the hot one down, the cold one up.
Check. The total change is zero, as it must be for a reversible engine.
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