The efficiency of a Carnot engine when the so | vedclass

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

Question

(A) The efficiency $(\eta)$ of a Carnot engine is defined as the ratio of the work done by the engine to the heat absorbed from the source.For a rever

Vedclass · Accepted Answer

$\frac{T_1 - T_2}{T_1}$

Vedclass · Answer

(A) The efficiency $(\eta)$ of a Carnot engine is defined as the ratio of the work done by the engine to the heat absorbed from the source.For a reversible Carnot cycle,the efficiency depends only on the temperatures of the source and the sink.The formula for the efficiency of a Carnot engine is given by:$\eta = 1 - \frac{T_2}{T_1} = \frac{T_1 - T_2}{T_1}$Where:$T_1$ is the absolute temperature of the source (in Kelvin).$T_2$ is the absolute temperature of the sink (in Kelvin).Therefore,the correct option is $A$.

The efficiency of a Carnot engine when the source temperature is $T_1$ and the sink temperature is $T_2$ is given by:

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 $1.0\,kW$?

An engine performs $2000 \,J$ of mechanical work and discards $4000 \,J$ of heat each cycle. What is the thermal efficiency of the engine (in $\%$)?

Which type of process is performed in an ideal engine?

The first operation involved in a Carnot cycle is

$A$ Carnot engine with efficiency $50\,\%$ takes heat from a source at $600\,K$. In order to increase the efficiency to $70\,\%$,keeping the temperature of the sink the same,the new temperature of the source will be $.........\,K$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module