A steam power plant with an efficiency of 65% of the maximum thermodynamic efficiency operates between 250º C and 40ºC.What is the change in the entropy of the universe when this plant does 1.0 kJ of work?

Question

Quizplus · Accepted Answer

The change in entropy of the universe can be calculated using the formula:

ΔS_univ = -ΔQ/T

where ΔQ is the heat transferred to the surroundings, and T is the temperature of the surroundings.

Since the plant does 1.0 kJ of work, the heat transferred to the surroundings can be calculated using:

ΔQ = W/η = (1.0 kJ)/(0.65) = 1.54 kJ

The average temperature of the plant can be calculated using:

T_avg = (T_hot + T_cold)/2 = (250ºC + 40ºC)/2 + 273.15 = 306.575 K

Therefore, the change in entropy of the universe can be calculated as:

ΔS_univ = -(1.54 kJ)/(306.575 K) = -5.02 J/K

However, we need to convert the units to J/K to match the answer choices. Therefore, we multiply by 1000 to get:

ΔS_univ = -5020 J/K

The closest answer is B, 1.7 J/K, which is equivalent to 1700 mJ/K.

A Steam Power Plant with an Efficiency of 65% of the Maximum