The efficiency of a Carnot heat engine is 0.5 | vedclass

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

Question

(A) The efficiency $\eta$ of a Carnot heat engine is given by the formula: $\eta = 1 - \frac{T_{2}}{T_{1}}$.Given that $\eta = 0.5$, we have $0.5 = 1

Vedclass · Accepted Answer

$2 \,T_{1}, 2 \,T_{2}$

Vedclass · Answer

(A) The efficiency $\eta$ of a Carnot heat engine is given by the formula: $\eta = 1 - \frac{T_{2}}{T_{1}}$.Given that $\eta = 0.5$, we have $0.5 = 1 - \frac{T_{2}}{T_{1}}$, which implies $\frac{T_{2}}{T_{1}} = 0.5$, or $T_{1} = 2T_{2}$.For another Carnot engine to have the same efficiency of $0.5$, the ratio of the sink temperature to the source temperature must remain the same, i.e., $\frac{T_{2}'}{T_{1}'} = 0.5$.If we multiply both the source and sink temperatures by the same factor $k$, the new efficiency is $\eta' = 1 - \frac{k T_{2}}{k T_{1}} = 1 - \frac{T_{2}}{T_{1}} = 0.5$.Therefore, if the source temperature is $2T_{1}$ and the sink temperature is $2T_{2}$, the efficiency remains $0.5$.

The efficiency of a Carnot heat engine is $0.5$ when the temperature of the source is $T_{1}$ and that of the sink is $T_{2}$. The efficiency of another Carnot heat engine is also $0.5$. The temperatures of the source and sink of the second engine are respectively:

Similar Questions

The efficiency of an ideal Carnot engine working between temperatures $T_1$ and $T_2$ is $1/3$. If the temperature of the sink is reduced by $40 \%$,then its efficiency will be: (in $\%$)

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

The efficiency of a Carnot engine is $50\%$ and the temperature of the sink is $500 \ K$. If the temperature of the source is kept constant and its efficiency is raised to $60\%$,then the required temperature of the sink will be: (in $K$)

$A$ Carnot engine has an efficiency of $40\%$. The temperature of its sink is $300 \ K$. To increase the efficiency by $50\%$ of its original efficiency while keeping the sink temperature constant,by how many $K$ must the source temperature be increased?

If the ratio of the absolute temperature of the sink and source of a Carnot engine is changed from $2:3$ to $3:4$,the efficiency of the engine changes by (in $\%$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module