/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 73 A Carnot engine takes \(3 \times... [FREE SOLUTION] | 91影视

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Chapter 12: Problem 73

A Carnot engine takes \(3 \times 10^{6}\) cal of heat from a reservoir at \(627^{\circ} \mathrm{C}\) and gives it to a sink at \(27^{\circ} \mathrm{C}\). The work done by the engine is [2003] (A) \(4.2 \times 10^{6} \mathrm{~J}\) (B) \(8.4 \times 10^{6} \mathrm{~J}\) (C) \(16.8 \times 10^{6} \mathrm{~J}\) (D) Zero

Short Answer

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

Step by step solution

01

Convert temperatures to Kelvin

The temperatures are given in Celsius, but we need to work with them in Kelvin. To convert a temperature from Celsius to Kelvin, add 273.15 to it. `High Temperature` \((T_H)\) = \(627^{\circ} \mathrm{C} + 273.15 = 900.15 \mathrm{~K}\) `Low Temperature` \((T_L)\)= \(27^{\circ} \mathrm{C} + 273.15 = 300.15 \mathrm{~K}\)
02

Calculate the efficiency of the Carnot engine

The efficiency (\(\eta\)) of a Carnot engine is given by: \(\eta = 1 - \frac{T_L}{T_H}\) Plugging in the values we found in Step 1: \(\eta = 1 - \frac{300.15}{900.15} = 1 - 0.333 = 0.667\)
03

Convert Heat to Joules

The heat taken from the reservoir is given in calories. We need it in Joules. To convert calories to Joules, use the following conversion: `1 cal = 4.186 J` Heat taken from reservoir in Joules, Q_H: \(Q_H = 3 \times 10^{6} \mathrm{~cal} \times 4.186 \mathrm{~J/cal} = 1.2558 \times 10^{7} \mathrm{~J}\)
04

Calculate work done by the engine

Now, we have the efficiency (\(\eta\)) of the Carnot engine and the heat taken from the reservoir \(Q_H\). We can calculate the work done by the engine, W, using the following formula: \(W = \eta \times Q_H\) Plugging in the values we found: \(W = 0.667 \times 1.2558 \times 10^{7} \mathrm{~J} \approx 8.4 \times 10^{6} \mathrm{~J}\) The work done by the engine is \(8.4 \times 10^{6} \mathrm{~J}\), which corresponds to option (B).

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Most popular questions from this chapter

The internal energy \(U\) is a unique function of any state because of change in \(U\) (A) does not depend upon path. (B) depends upon path. (C) corresponds to an adiabatic process. (D) corresponds to an isothermal process.

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An ideal gas undergoes a quasi-static, reversible process in which its molar heat capacity \(C\) remains constant. If during this process the relation of pressure \(P\) and volume \(V\) is given by \(P V^{n}=\) constant, then \(n\) is given by (Here \(C_{P}\) and \(C_{V}\) are molar specific heat of constant pressure and constant volume, respectively) [2016] (A) \(n=\frac{C-C_{P}}{C-C_{V}}\) (B) \(n=\frac{C_{P}-C}{C-C_{V}}\) (C) \(n=\frac{C-C_{V}}{C-C_{P}}\) (D) \(n=\frac{C_{P}}{C_{V}}\)

Two moles of an ideal gas at \(300 \mathrm{~K}\) were cooled at constant volume so that the pressure is reduced to half the initial value. As a result of heating at constant pressure, the gas has expanded till it attains the original temperature. The total heat absorbed by gas, if \(R\) is the gas constant (A) \(150 R\) (B) \(300 R\) (C) \(75 R\) (D) \(100 R\)

One mole of an ideal monatomic gas is taken from temperature \(T_{0}\) to \(2 T_{0}\) by the process \(P T^{-4}=C\) Considering the following statements, choose the correct alternative. I. Molar heat capacity of the gas is \(-\frac{3 R}{2}\) II. Molar heat capacity of the gas is \(\frac{3 R}{2}\) III. Work done is \(-3 R T_{0}\) IV. Work done is \(3 R T_{0}\) (A) Statements I and IV are correct. (B) Statements I and III are correct. (C) Statements II and IV are correct. (D) Statements II and III are correct.
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