Question

A heat engine is involved with the exchange of heat of $1950J$, $-40J$, $+125J$ and $-QJ$, during one cycle achieving an efficiency of $50.0\%$. the value of $Q$ is:

• A. $980J$
• B. $640J$
• C. $40J$
• D. $400J$

Answer 1

The correct option is A

$980J$

Explanation for the correct option:

A heat engine is a device that converts heat energy into mechanical work through a cyclic process.

The efficiency of a heat engine is the ratio of work done by the heat engine to the energy absorbed per cycle.

It is given that the efficiency of heat engine, $\mathrm{\eta }=\frac{1}{2}$, heat released by the engine in doing work, $Q_R=-40Jand-QJ$ and the heat supplied, $Q_S=1915Jand125J$.

The formula for efficiency of heat engine is: $\mathrm{\eta }=1-\frac{{\mathrm{Q}}_{\mathrm{R}}}{{\mathrm{Q}}_{\mathrm{S}}}$ Where, $\mathrm{\eta }$is the efficiency of heat engine, ${\mathrm{Q}}_{\mathrm{R}}$is heat released and ${\mathrm{Q}}_{\mathrm{S}}$ is the heat supplied.

Putting the values in the formula we get : $\frac{1}{2}=1-\frac{\left(40+Q\right)}{\left(1915+125\right)}$

$\frac{40+Q}{2040}=1-\frac{1}{2}$

$80+2Q=2040$

$$\begin{gathered} 2Q=2040-80 \\ Q=\frac{1960}{2} \\ Q=980 \end{gathered}$$

Hence, $\mathbf{Q}\mathbf{=}\mathbf{980}\mathbf{}\mathbf{J}$.

Thus option A is correct.