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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

A carnot's engine whose sink is at a temperature of \(300 \mathrm{~K}\) has an efficiency of \(40 \%\) By what amount should the temperature of the source change to increase the efficiency to \(60 \%\) (A) \(275 \mathrm{~K}\) (B) \(325 \mathrm{~K}\) (C) \(300 \mathrm{~K}\) (D) \(250 \mathrm{~K}\)

If heat given to a system is \(6 \mathrm{k}\) cal and work done is $6 \mathrm{kj}$. The change in internal energy is ......... KJ. (A) \(12.4\) (B) 25 (C) \(19.1\) (D) 0

In Column I different Process is given match corresponding option of column \(\mathrm{I}_{1}\) Column - I Columm - II (A) adiabatic process (p) \(\Delta \mathrm{p}=0\) (B) Isobaric process (q) \(\Delta \mathrm{u}=0\) (C) Isochroic process (r) \(\Delta Q=0\) (D) Isothermal process (s) \(\Delta \mathrm{W}=0\) (A) A-p, B-s, C-r, D-q (B) (B) A-s, B-q, C-p, D-r (C) $\mathrm{A}-\mathrm{r}, \mathrm{B}-\mathrm{p}, \mathrm{C}-\mathrm{s}, \mathrm{D}-\mathrm{q}$ (D) $\mathrm{A}-\mathrm{q}, \mathrm{B}-\mathrm{r}, \mathrm{C}-\mathrm{q}, \mathrm{D}-\mathrm{p}$

An engine is supposed to operate between two reservoirs at temperature \(727^{\circ} \mathrm{C}\) and \(227^{\circ} \mathrm{C}\). The maximum possible efficiency of such an engine is (A) \((3 / 4)\) (B) \((1 / 4)\) (C) \((1 / 2)\) (D) 1

The temperature on Celsius scale is \(25^{\circ} \mathrm{C}\). What is the corresponding temperature on the Fahrenheit Scale? (A) \(40^{\circ} \mathrm{F}\) (B) \(45^{\circ} \mathrm{F}\) (C) \(50^{\circ} \mathrm{F}\) (D) \(77^{\circ} \mathrm{F}\)
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