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Chapter 8: Problem 1200

If a heat engine absorbs \(50 \mathrm{KJ}\) heat from a heat source and has efficiency of \(40 \%\), then the heat released by it in heat sink is (A) \(40 \mathrm{KJ}\) (B) \(30 \mathrm{KJ}\) (C) \(20 \mathrm{~J}\) (D) \(20 \mathrm{KJ}\)

Short Answer

Expert verified
The heat released by the heat engine in the heat sink is 30 KJ. Therefore, the correct answer is option (B) 30 KJ.

Step by step solution

01

Write given information

We are given: - Heat absorbed by the engine (Q_in) = 50 KJ - Efficiency (η) = 40%
02

Calculate work done by engine

Efficiency (η) = work done (W) / heat absorbed (Q_in) We will rearrange to find the work done and convert efficiency to a decimal: W = η * Q_in W = 0.4 * 50 KJ
03

Calculate heat released by the engine

Since energy is conserved, the heat absorbed by the engine must be equal to the sum of the work done and the heat released by the engine in the heat sink (Q_out): Q_in = W + Q_out Now we'll rearrange the equation to find Q_out and insert the values found in the previous steps: Q_out = Q_in - W Q_out = 50 KJ - (0.4 * 50 KJ)
04

Evaluate heat released by the engine

After solving the equation for Q_out, we get: Q_out = 50 KJ - (0.4 * 50 KJ) = 50 KJ - 20 KJ = 30 KJ
05

State the final answer

The heat released by the heat engine in the heat sink is 30 KJ. Therefore, the correct answer is option (B) 30 KJ.

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Most popular questions from this chapter

One mole of an ideal gas $\left(\mathrm{C}_{\mathrm{p}} / \mathrm{C}_{\mathrm{v}}\right)=\gamma$ at absolute temperature \(\mathrm{T}_{1}\) is adiabatically compressed from an initial pressure \(\mathrm{P}_{1}\) to a final pressure \(\mathrm{P}_{2}\) The resulting temperature \(\mathrm{T}_{2}\) of the gas is given by. (A) $\mathrm{T}_{2}=\mathrm{T}_{1}\left\\{\mathrm{p}_{2} / \mathrm{p}_{1}\right\\}^{\\{\gamma /(\gamma-1)\\}}$ (B) $\mathrm{T}_{2}=\mathrm{T}_{1}\left\\{\mathrm{p}_{2} / \mathrm{p}_{1}\right\\}^{\\{(\gamma-1) / \gamma\\}}$ (C) $\mathrm{T}_{2}=\mathrm{T}_{1}\left\\{\mathrm{p}_{2} / \mathrm{p}_{1}\right\\}^{\gamma}$ (D) $\mathrm{T}_{2}=\mathrm{T}_{1}\left(\mathrm{p}_{2} / \mathrm{p}_{1}\right)^{\gamma-1}$

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