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

option (d)

Step-by-step 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-(Q_(c))/(Q_(H))`

`0.55=1-\frac{800}}{Q_(H)}`

QH = 1777.77 J

`\Delta S_(1)=-(Q_(H))/(T_(H))=(-1777.77)/(500)=-3.55 J/K`

Where, ΔS1 is the change in entropy

`\Delta S_(2)=(Q_(C))/(T_(C))=(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)