Determine the efficiency of an ideal heat engine with a heater temperature of 480 degrees

Determine the efficiency of an ideal heat engine with a heater temperature of 480 degrees and a refrigerator temperature of minus 30 degrees.

Given: T1 (temperature that the heater has) = 480 ºС; T2 (temperature that the refrigerator has) = -30 ºС.

1) Let’s translate the set temperatures from the Celsius scale to the Kelvin scale:

a) Т1 = 480 ºС = 480 + 273 К = 753 К;

b) Т2 = -30 ºС = -30 + 273 К = 243 K.

2) Calculate the efficiency: η = ((T1 – T2) / T2) * 100% = ((753 – 243) / 753) * 100% = 0.677 * 100% = 67.7%.

Answer: An ideal heat engine will have an efficiency of 67.7%.