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During the isothermal heat rejection process of a Carnot cycle, the working fluid experiences an entropy change of -0.7 Btu/R. If the temperature of the heat sink is 95 degree F, determine (a) the amount of heat transfer, (b) the entropy change of the sink, and (c) the total entropy change for this process.

User Atir
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Final answer:

During the isothermal heat rejection process of a Carnot cycle, the amount of heat transfer is -66.5 Btu, the entropy change of the sink is -66.5 Btu/R, and the total entropy change for this process is -67.2 Btu/R.

Step-by-step explanation:

The amount of heat transfer for the isothermal heat rejection process in a Carnot cycle can be determined using the formula:

Q = T * ΔS

Where Q is the amount of heat transfer, T is the temperature of the heat sink, and ΔS is the change in entropy. Plugging in the given values, we have:

Q = 95 × (-0.7)

Simplifying, we get:

Q = -66.5 Btu

The negative sign indicates that heat is being rejected from the working fluid.

The entropy change of the heat sink can be determined using the same formula, but with the entropy change being positive since heat is being absorbed:

ΔSsink = Tsink * ΔS

Plugging in the given values, we have:

ΔSsink = 95 * (-0.7)

Simplifying, we get:

ΔSsink = -66.5 Btu/R

The total entropy change for this process is the sum of the entropy change of the working fluid and the entropy change of the sink:

Total ΔS = ΔSworking fluid + ΔSsink

Plugging in the given values, we have:

Total ΔS = -0.7 + (-66.5)

Simplifying, we get:

Total ΔS = -67.2 Btu/R

User J Pollack
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