Question

15. A heat engine operating between temperatures of 100 K and 500 K has an efficiency of 55%. If each cycle expels 800 J of heat, what is the change in entropy of the universe due to each cycle? a) 5.33 J/K b) 5.09 J/K c) 4.80 J/K d) 4.44 J/K e) 0.00 J/K

Answer 1

Answer:

option (d)

Explanation:

temperature at cold bath, Tc = 100 K

temperature at hot bath, TH = 500 K

Efficiency = 55 %

heat expelled, Qc = 800 J

Let the heat intake is QH.

Use the formula for the efficiency

$\eta =1-\frac{Q_{c}}{Q_{H}}$

$0.55=1-\frac{800}}{Q_{H}}$

QH = 1777.77 J

$\Delta S_{1}=-\frac{Q_{H}}{T_{H}}=\frac{-1777.77}{500}=-3.55 J/K$

Where, ΔS1 is the change in entropy

$\Delta S_{2}=\frac{Q_{C}}{T_{C}}=\frac{800}{100}=8 J/K$

$\Delta S_{universe}=\Delta S_{1}+\Delta S_{2}$

ΔS(universe) = 8 - 3.55 = 4.44 J/K

option (d)