A Carnot engine takes 3 times 10^6 calories o | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Heat taken by the Carnot engine,$Q = 3 	imes 10^6 	ext{ cal} = 4.2 	imes 3 	imes 10^6 	ext{ J} = 12.6 	imes 10^6 	ext{ J}$.Temperature of t

Vedclass · Accepted Answer

$8.4 	imes 10^6 	ext{ J}$

Vedclass · Answer

(B) Heat taken by the Carnot engine,$Q = 3 	imes 10^6 	ext{ cal} = 4.2 	imes 3 	imes 10^6 	ext{ J} = 12.6 	imes 10^6 	ext{ J}$.Temperature of the source,$T_1 = (627 + 273) 	ext{ K} = 900 	ext{ K}$.Temperature of the sink,$T_2 = (27 + 273) 	ext{ K} = 300 	ext{ K}$.Efficiency of a Carnot engine is given by $\eta = \frac{W}{Q} = 1 - \frac{T_2}{T_1}$.Substituting the values: $\frac{W}{12.6 	imes 10^6} = 1 - \frac{300}{900}$.$\frac{W}{12.6 	imes 10^6} = 1 - \frac{1}{3} = \frac{2}{3}$.$W = \frac{2}{3} 	imes 12.6 	imes 10^6 	ext{ J} = 8.4 	imes 10^6 	ext{ J}$.

$A$ Carnot engine takes $3 \times 10^6$ calories of heat from a reservoir at $627^{\circ} C$ and gives it to a sink at $27^{\circ} C$. The work done by the engine is

Similar Questions

The efficiency of a Carnot engine is $100\%$ if

$A$ reversible engine converts one-sixth of the heat input into work. When the temperature of the sink is reduced by $62^\circ C$,the efficiency of the engine is doubled. The temperatures of the source and sink are:

In a Carnot engine,the temperature of the reservoir is $927\ ^oC$ and that of the sink is $27\ ^oC.$ If the work done by the engine is $12.6 \times 10^6 \ J,$ the quantity of heat absorbed by the engine from the reservoir is:

$A$ Carnot engine is made to work between $200^{\circ}C$ and $0^{\circ}C$ first and then between $0^{\circ}C$ and $-200^{\circ}C$. The ratio of efficiencies of the engine in the two cases is

The efficiency of a Carnot engine operating between reservoirs maintained at temperatures $27^{\circ}C$ and $-123^{\circ}C$ is ...... $\%$

Vedclass Test Series

Exam Paper Generator

Online Exam Module