Question

How much heat is exhausted to the cold reservoir by the heat engine shown in $FIGUREEX21.16$?

Short answer

The heat exhausted to the cold reservoir by heat engine,$Q_c=285J.$

Step-by-step solution

Step: 1 Definition of heat transfer in cold reservoirs:

(a) Heat moves spontaneously from a hot to a cold item, according to the second rule of thermodynamics.

(b) A heat engine, shown as a circle, operates by utilising some of the heat transmission. The hot and cold things are the heat and cold reservoirs.

Step: 2 Finding the value of Qh:

We can find the lost heat as

$Q_h=W+Q_c\Rightarrow Q_c=Q_h-W$

Heat taken from the heat reservoir,$Q_h$,will be equal tothe sum of the heat taken in the two processes.That is,

$Q_h=90+225=315\mathrm{J}.$

Step: 3 Finding the value of Qc:

The work finish on the gas will be same to the space encircled by the graph. Since this is the triangle, we have

$$\begin{gathered} W=\frac{1}{2}\cdot (300-100)\cdot {10}^3\cdot (600-300)\cdot {10}^{-6} \\ W=30\mathrm{J}. \end{gathered}$$

Finally, the heat lost to the cold reservoir is

$$\begin{gathered} Q_c=315-30 \\ {Q}_c=285\mathrm{J}. \end{gathered}$$