Question

Efficiency of a cyclic process is $50\%$ and heat interactions are $Q_1=+1915\text{J},Q_2=-40\text{J},Q_3=+125\text{J}$ and $Q_4=-Q\text{J}$. Find the value of $Q$.

• A. $1080\text{J}$
• B. $1980\text{J}$
• C. $-1080\text{J}$
• D. $980\text{J}$

Answer 1

The correct option is D $980\text{J}$

We know that efficiency$\left(\eta \right)$ of a cyclic process is given by

$\eta =\frac{\text{work done}\left(W\right)}{\text{Heat absorbed per cycle}}...\left(1\right)$

Here, heat absorbed,

$Q_{ab}=Q_1+Q_3=1915+125$

$\Rightarrow Q_{ab}=2040\text{J}.......\left(2\right)$

From first law of thermodynamics, work done $\left(W\right)$ for a cyclic process is given by

$W=\text{total heat absorbed}-\text{total heat rejected}$

$\Rightarrow W=(Q_1+Q_3)-(Q_2+Q_4)$

$\Rightarrow W=(1915+125)-(40+Q)$

$\Rightarrow W=2000-Q.......\left(3\right)$

From Eqs. (1), (2) and (3),

$\eta =\frac{2000-Q}{2040}$

$\Rightarrow 0.5=\frac{2000-Q}{2040}$

$\Rightarrow Q=980\text{J}$

So, $\left(d\right)$ is the correct answer.