What is the Hot and cold reservoir in this Carnot engine problem?

Question

The hot-reservoir temperature of a Carnot engine with 29.0% efficiency is 80.0 degrees C higher than the cold-reservoir temperature. What are the reservoir temperatures, in Celcius?

What i have so far :

n = 29% = .29

change in temp = 80 C = 353.15 K

.29 Th = 353.15 K

Th = 353.15/.29 = 1217 K = 944C

Th - Tc = 353.15

Tc = Th - 353.15 = 1217.75 K - 353.15K = 864.61 K = 591 C

My answer is still incorrect?

Can someone help me please?

Answer 1

n = 1 - Tc/Th = Th/Th - Tc/Th

thus n * Th = Th - Tc

and Th = (Th - Tc)/n

so Th = 80C° / 0.29 = 275.862°K

Tc = 195.86°K

check: n = 1 - 196/275 = 0.29

Your problem is that converting the 80C° difference in temp to K° is incorrect. The reason is that when you convert the temp of the resevoirs to °C, and subtract, you will see they are still 80C° (or K°, same thing) apart. Thus, 80 K°= 80C°; but 80°K = 353°C.

I was taught to reverse the degree symbol and letter to denote delta T, to help keep them straight. Not many people seem to do this anymore.

Answer 2

80 C is the difference in temperature, not the temperature of a source. The difference in temperature has the same measure in Celcius and Kelvin scale (you can prove it).

Therefore Th - Tc = 80 C = 80 K

n = (Th - Tc) / Th

Th = (Th - Tc) / n

Th = 80 / 0.29 = 276 K

The temperature of hot source is 276 K or 3 C

The temperature of cold source is  196 K or -77 C

Is this engine a refrigerator?