A Carnot engine absorbs an amount Q of heat f | vedclass

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

Question

(B) For a Carnot engine,the efficiency $\eta$ is given by the ratio of temperatures of the sink $(T_2)$ and the source $(T_1)$:$\eta = 1 - \frac{T_2}{

Vedclass · Accepted Answer

$Q / 3$

Vedclass · Answer

(B) For a Carnot engine,the efficiency $\eta$ is given by the ratio of temperatures of the sink $(T_2)$ and the source $(T_1)$:$\eta = 1 - \frac{T_2}{T_1}$Given,$T_1 = T$ and $T_2 = T/3$.Therefore,$\eta = 1 - \frac{T/3}{T} = 1 - \frac{1}{3} = \frac{2}{3}$.Also,the efficiency is defined as $\eta = 1 - \frac{Q_2}{Q_1}$,where $Q_1$ is the heat absorbed and $Q_2$ is the heat rejected.Equating the two expressions for efficiency:$\frac{2}{3} = 1 - \frac{Q_2}{Q}$$\frac{Q_2}{Q} = 1 - \frac{2}{3} = \frac{1}{3}$$Q_2 = \frac{Q}{3}$.Thus,the amount of heat rejected is $Q/3$.

$A$ Carnot engine absorbs an amount $Q$ of heat from a reservoir at an absolute temperature $T$ and rejects heat to a sink at a temperature of $T/3$. The amount of heat rejected is

Similar Questions

$A$ Carnot's heat engine works between the temperatures $427^{\circ} C$ and $27^{\circ} C$. How much heat in $kcal/s$ should it consume per second to deliver mechanical work at the rate of $1.0\,kW$?

For an ideal heat engine,the temperature of the source is $127\,^{\circ} C$. In order to have $60\, \%$ efficiency,the temperature of the sink should be $........\,{ }^{\circ} C$. (Round off to the nearest integer)

$A$ Carnot engine working between $400 \, K$ and $800 \, K$ has a work output of $1200 \, J$ per cycle. What is the amount of heat energy supplied to the engine from the source per cycle in $J$?

What is the theoretical efficiency of a heat engine?

Vedclass Test Series

Exam Paper Generator

Online Exam Module