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Chapter 8: Problem 1201

The efficiency of heat engine is \(30 \%\) If it gives \(30 \mathrm{KJ}\) heat to the heat sink, than it should have absorbed ....... KJ heat from heat source. (A) \(42.8\) (B) 39 (C) 29 (D) 9

Short Answer

Expert verified
The heat engine should have absorbed approximately 42.8 KJ of heat from the heat source. The correct answer is (A).

Step by step solution

01

Write down the given information

We are given: - Efficiency of heat engine = 30% - Heat given to the heat sink = 30 KJ
02

Calculate work done by engine

Using the efficiency formula, we can calculate the work done by the engine: Efficiency = \(\frac{Work\_done\_by\_engine}{Heat\_absorbed\_from\_heat\_source}\) Rearrange the equation to find the work done by the engine: Work\_done\_by\_engine = Efficiency × Heat\_absorbed\_from\_heat\_source Since we don't know the heat absorbed from the heat source yet, let's represent it with the variable H: Work\_done\_by\_engine = 0.3 × H
03

Write the equation for heat absorbed from heat source

The heat absorbed from the heat source is equal to the sum of the heat given to the heat sink and the work done by the engine: Heat\_absorbed\_from\_heat\_source = Heat\_given\_to\_heat\_sink + Work\_done\_by\_engine Plug in the given values and the equation we obtained in Step 2: H = 30 + (0.3 × H)
04

Solve for H

Now we can solve the equation for H: 0.7 × H = 30 H = \(\frac{30}{0.7}\) H = 42.8571428571 KJ
05

Compare the result to the answer choices

Comparing our result to the answer choices, we find that: (A) 42.8 KJ is the closest to our calculated heat absorbed value. Therefore, the heat engine should have absorbed approximately 42.8 KJ of heat from the heat source. The correct answer is (A).

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Most popular questions from this chapter

When a System is taken from State \(i\) to State \(f\) along the path iaf, it is found that \(\mathrm{Q}=70 \mathrm{cal}\) and \(\mathrm{w}=30 \mathrm{cal}\), along the path ibf. \(\mathrm{Q}=52\) cal. \(\mathrm{W}\) along the path ibf is (A) 6 cal (B) \(12 \mathrm{cal}\) (C) \(24 \mathrm{cal}\) (D) 8 cal

Work done per mole in an isothermal ? change is (A) RT log \(_{e}\left(\mathrm{v}_{2} / \mathrm{v}_{1}\right)\) (B) \(\mathrm{RT} \log _{10}\left(\mathrm{v}_{2} / \mathrm{v}_{1}\right)\) (C) RT \(\log _{10}\left(\mathrm{v}_{1} / \mathrm{v}_{2}\right)\) (D) RT \(\log _{\mathrm{e}}\left(\mathrm{v}_{1} / \mathrm{v}_{2}\right)\)

A Carnot engine Converts one sixth of the heat input into work. When the temperature of the sink is reduced by \(62^{\circ} \mathrm{C}\) the efficiency of the engine is doubled. The temperature of the source and sink are (A) \(80^{\circ} \mathrm{C}, 37^{\circ} \mathrm{C}\) (B) \(95^{\circ} \mathrm{C}, 28^{\circ} \mathrm{C}\) (C) \(90^{\circ} \mathrm{C}, 37^{\circ} \mathrm{C}\) (D) \(99^{\circ} \mathrm{C}, 37^{\circ} \mathrm{C}\)

For an adiabatic process involving an ideal gas (A) \(\mathrm{P}^{\gamma-1}=\mathrm{T}^{\gamma-1}=\) constant (B) \(\mathrm{P}^{1-\gamma}=\mathrm{T}^{\gamma}=\) constant (C) \(\mathrm{PT}^{\gamma-1}=\) constant (D) \(\mathrm{P}^{\gamma-1}=\mathrm{T}^{\gamma}=\) constant

The change in internal energy, when a gas is cooled from $927^{\circ} \mathrm{C}\( to \)27^{\circ} \mathrm{C}$ (A) \(200 \%\) (B) \(100 \%\) (C) \(300 \%\) (D) \(400 \%\)
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