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Chapter 8: Problem 1189

A Carnot engine takes \(3 \times 10^{6}\) cal of heat from a reservoir at \(627^{\circ} \mathrm{C}\), and gives to a sink at \(27^{\circ} \mathrm{C}\). The work done by the engine is (A) \(4.2 \times 10^{6} \mathrm{~J}\) (B) \(16.8 \times 10^{6} \mathrm{~J}\) (C) \(8.4 \times 10^{6} \mathrm{~J}\) (D) Zero

Short Answer

Expert verified
The work done by the Carnot engine is approximately \(8.4 \times 10^{6} \mathrm{~J}\) (Option C).

Step by step solution

01

Convert temperatures to Kelvin

To work with temperatures in thermodynamics, we need to convert them to Kelvin scale. The conversion formula is: K = °C + 273.15 For the heat source: \(T_1 = 627^\circ C + 273.15 = 900.15 K\) For the heat sink: \(T_2 = 27^\circ C + 273.15 = 300.15 K\)
02

Calculate the efficiency of the Carnot engine

The efficiency of a Carnot engine is given by the formula: \(\eta = 1 - \frac{T_2}{T_1}\) Substitute the values of \(T_1\) and \(T_2\) into the formula: \(\eta = 1 - \frac{300.15}{900.15} = 1 - 0.33333 = 0.66667\)
03

Calculate the work done by the engine

We know the heat input \(Q_1\) to the engine is \(3 \times 10^6\) cal. We need to convert this to Joules. The conversion factor is 1 cal = 4.184 J. Thus, \(Q_1 = 3 \times 10^6 \mathrm{~cal} \times 4.184 \mathrm{~J/cal} = 1.2552 \times 10^7 \mathrm{~J}\) Now, we can use the efficiency to find the work done by the engine. The efficiency is equal to the ratio of work done \(W\) to the heat input \(Q_1\): \(\eta = \frac{W}{Q_1}\) Rearranging for W and substituting the values for \(\eta\) and \(Q_1\): \(W = \eta \times Q_1 = 0.66667 \times 1.2552 \times 10^7 \mathrm{~J} = 8.37 \times 10^6 \mathrm{~J}\) The closest option is (C) \(8.4 \times 10^{6} \mathrm{~J}\).

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Most popular questions from this chapter

The latent heat of Vaporization of water is \(2240(\mathrm{~J} / \mathrm{g})\) If the work done in the Process of expansion of \(1 \mathrm{~g}\) is $168 \mathrm{~J}$. then increase in internal energy is ......... J (A) 2072 (B) 2408 (C) 2240 (D) 1904

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