Question

A completely reversible heat engine operates with a source at 1500 R and a sink at 500 R. If the entropy of the sink increases by 10 Btu/R, how much will the entropy of the source decrease? How much heat, in Btu, is transferred from the source?

Answer 1

The entropy of the source will decrease by the same amount as the entropy increase of the sink. The heat transferred from the source to the sink is 5000 Btu.

Step-by-step explanation:

In a completely reversible heat engine, the entropy change of the source and the sink are equal in magnitude but opposite in sign. So, if the entropy of the sink increases by 10 Btu/R, the entropy of the source will decrease by the same amount.

The heat transferred from the source to the sink can be calculated using the entropy change formula: ΔS = Q/T, where ΔS is the change in entropy, Q is the heat transfer, and T is the temperature. In this case, the sink temperature is 500 R and the entropy change is 10 Btu/R. Plugging these values into the formula, we can find the heat transfer: 10 Btu/R = Q/500 R. Rearranging the equation, we get Q = 5000 Btu.

Answer 2

Step-by-step explanation:

Efficiency of reversible engine

= (T₁ - T₂) / T₁ , T₁ and T₂ be temperature of source and sink respectively

= (1500 - 500 )R / 1500R

= 1000 / 1500

= 2 / 3

If Q₁ be heat extracted and Q₂ be heat dumped into sink

efficiency = (Q₁ - Q₂) / Q₁

given Q₂ / T₂ = 10

Q₂ = 10 T₂ = 10 x 500R = 5000BTu

Q₁ - 5000R / Q₁ = 2/3

3Q₁ - 15000R = 2Q₁

Q₁ = 15000BTu

Q₂ = 5000 BTu

Q₁ / T₁ = 15000R/ 1500R

= 10R

Decrease in entropy of source = 10R.

Heat transferred = Q₁ - Q₂

= 15000 - 5000 BTu

= 10000BTu

=