An ideal heat engine operates in a Carnot cyc | vedclass

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

Question

(C) For a Carnot cycle,the efficiency $\eta$ is given by $\eta = 1 - \frac{T_2}{T_1} = \frac{W}{Q_1}$.Given temperatures in Kelvin: $T_1 = 127 + 273 =

Vedclass · Accepted Answer

$1.25 	imes 10^4 	ext{ cal}$

Vedclass · Answer

(C) For a Carnot cycle,the efficiency $\eta$ is given by $\eta = 1 - \frac{T_2}{T_1} = \frac{W}{Q_1}$.Given temperatures in Kelvin: $T_1 = 127 + 273 = 400 	ext{ K}$ and $T_2 = 27 + 273 = 300 	ext{ K}$.Given heat absorbed $Q_1 = 5 	imes 10^4 	ext{ cal}$.Substituting the values into the efficiency formula:$\eta = 1 - \frac{300}{400} = 1 - 0.75 = 0.25$.Since $\eta = \frac{W}{Q_1}$,we have $W = \eta 	imes Q_1$.$W = 0.25 	imes 5 	imes 10^4 	ext{ cal} = 1.25 	imes 10^4 	ext{ cal}$.Thus,the amount of heat converted to work is $1.25 	imes 10^4 	ext{ cal}$.

An ideal heat engine operates in a Carnot cycle between $127^{\circ} C$ and $27^{\circ} C$. It absorbs $5 \times 10^4 \text{ cal}$ of heat at the higher temperature. The amount of heat converted to work is:

Similar Questions

$A$ Carnot engine absorbs $3 \times 10^{6} \text{ cal}$ of heat from a reservoir at $627^{\circ}C$ and rejects it at $27^{\circ}C$. The work done by the engine is:

$A$ Carnot engine takes $5000 \, kcal$ of heat from a reservoir at $727 \, ^{\circ}C$ and gives heat to a sink at $127 \, ^{\circ}C$. The work done by the engine is $.......... \times 10^{6} \, J$.

An engine has an efficiency of $1/6$. When the temperature of the sink is reduced by $62^{\circ}C$,its efficiency is doubled. The temperature of the source is ....... $^{\circ}C$.

The efficiency of a Carnot engine is found to increase from $25 \%$ to $40 \%$ on increasing the temperature $(T_1)$ of the source alone by $100 \ K$. The temperature $(T_2)$ of the sink is given by: (in $K$)

$A$ Carnot engine converts $1/6$th of the heat into work. When the temperature of the sink is reduced by $62 \ K$,the efficiency of the engine is doubled. Calculate the temperatures of the source and the sink.

Vedclass Test Series

Exam Paper Generator

Online Exam Module