A heat engine is operating between 500 K and | vedclass

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

Question

(D) For a heat engine operating between temperatures $T_H = 500 \ K$ and $T_L = 300 \ K$,the efficiency $\eta$ is given by $\eta = 1 - \frac{T_L}{T_H}

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(D) For a heat engine operating between temperatures $T_H = 500 \ K$ and $T_L = 300 \ K$,the efficiency $\eta$ is given by $\eta = 1 - \frac{T_L}{T_H} = 1 - \frac{300}{500} = 1 - 0.6 = 0.4$.The efficiency is also defined as $\eta = 1 - \frac{Q_L}{Q_H}$,where $Q_H = 100 \ J$ is the heat absorbed and $Q_L$ is the heat rejected.According to the Second Law of Thermodynamics,the efficiency of any heat engine cannot exceed the Carnot efficiency: $1 - \frac{Q_L}{Q_H} \leq 1 - \frac{T_L}{T_H}$.This simplifies to $\frac{Q_L}{Q_H} \geq \frac{T_L}{T_H}$.Substituting the values: $\frac{Q_L}{100} \geq \frac{300}{500} = 0.6$.Therefore,$Q_L \geq 60 \ J$.This means the heat rejected must be at least $60 \ J$. Any value less than $60 \ J$ is impossible. Among the given options,$20 \ J$ is less than $60 \ J$,making it an impossible amount of heat rejected.

$A$ heat engine is operating between $500 \ K$ and $300 \ K$. If the engine absorbs $100 \ J$ of heat,which of the following is an impossible amount of heat rejected by the engine in $J$?

Similar Questions

Which of the following is correct regarding entropy?

Entropy is maximum in case of

The enthalpy of fusion of water is $6.01 \, kJ \, mol^{-1}$. The entropy change of $1 \, mole$ of ice at its melting point will be.....$J \, K^{-1} \, mol^{-1}$.

Calculate the entropy change for the melting of $1 \, kg$ of ice at $0 \, ^\circ C$ in $SI$ units. (Latent heat of fusion of ice = $80 \, cal \, g^{-1}$) (in $.67$)

$A$ reversible adiabatic process is one in which:

Vedclass Test Series

Exam Paper Generator

Online Exam Module