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,

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?

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Kiệt Gia 3 years 2021-07-27T11:12:03+00:00 1 Answers 88 views 0

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  1. quangkhai
    0
    2021-07-27T11:13:22+00:00
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    Answer:

    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

    =

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