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?

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

=

answer: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg