Question

An ideal Carnot engine operates between $500°\mathrm{C}$and $100°\mathrm{C}$with a heat input of 250 J per cycle. What minimum number of cycles is necessary for the engine to lift a $500\mathrm{kg}$rock through a height of $100\mathrm{m}$?

Short answer

The minimum number of cycles is necessary for the engine to lift a $500\mathrm{kg}$rock through a height of $100\mathrm{m}$is $3800\mathrm{cycles}$.

Step-by-step solution

Step 1:  Relation between temperature of reservoir, temperature of sink and heat gain  and heat reject by Carnot engine .

$\frac{{\mathrm{Q}}_2}{{\mathrm{Q}}_1}=\frac{{\mathrm{T}}_2}{{\mathrm{T}}_1}$

Where, ${\mathrm{T}}_1$is the temperature of reservoir, ${\mathrm{T}}_2$is the temperature of sink, ${\mathrm{Q}}_1$is the heat gain and ${\mathrm{Q}}_2$heat reject by Carnot engine.

Step 2:  Work done to lift the rock to a particular height

Work done in lifting the rock of mass $\mathrm{m}$to height $\mathrm{h}$ against gravity is given by

Work done = Force applied by external $\times$displacement

$\mathrm{W}=\mathrm{mgh}$

Work done by Carnot engine in one cycle

The total amount of heat gain by Carnot engine is equal to heat spends on useful work and rejected heat.

$$\begin{gathered} {\mathrm{Q}}_1=\mathrm{W}+{\mathrm{Q}}_2 \\ \mathrm{W}={\mathrm{Q}}_1-{\mathrm{Q}}_2 \end{gathered}$$

Calculation of heat rejected by Carnot engine

Using relation

$\frac{{\mathrm{Q}}_2}{{\mathrm{Q}}_1}=\frac{{\mathrm{T}}_2}{{\mathrm{T}}_1}$

Now , putting the values of constants in above

$$\begin{gathered} \frac{{\mathrm{Q}}_2}{250\mathrm{J}}=\frac{373\mathrm{K}}{773\mathrm{K}} \\ {\mathrm{Q}}_2=\left(250\right)\frac{373}{773} \\ {\mathrm{Q}}_2=121.0\mathrm{J} \end{gathered}$$

Here, ${\mathrm{Q}}_2$is the rejected by Carnot engine therefore, conventionally it must be negative.

So, ${\mathrm{Q}}_2=-121.0\mathrm{J}$

Calculation of work done by Carnot engine in one cycle

Using relation

${\mathrm{W}}_{\mathrm{E}}={\mathrm{Q}}_1-{\mathrm{Q}}_2$

Now, putting the values of constants in above equation

$$\begin{gathered} {\mathrm{W}}_{\mathrm{E}}=250.0\mathrm{J}-121.0\mathrm{J} \\ {\mathrm{W}}_{\mathrm{E}}=129.0\mathrm{J} \end{gathered}$$

Calculation of minimum number of cycles is necessary for the engine to lift a 500 kg rock through a height of 100 m

n minimum no of cycles$\times$Work done by engine in one cycle = Work done in lifting the stone .

$$\begin{gathered} \mathrm{n}\times {\mathrm{W}}_{\mathrm{E}}=\mathrm{W} \\ \mathrm{n}\times {\mathrm{W}}_{\mathrm{E}}=\mathrm{mgh} \end{gathered}$$

Now, putting the values of constants in above equation

$$\begin{gathered} \mathrm{n}\times 129.0\mathrm{J}=500.0\mathrm{kg}\times 9.8{\mathrm{ms}}^{-1}\times 100.0\mathrm{m} \\ \mathrm{n}=\frac{490.0\times {10}^3}{129.0} \\ \mathrm{n}=3800\mathrm{Cycles} \end{gathered}$$

Thus, The minimum number of cycles is necessary for the engine to lift a $500\mathrm{kg}$rock through a height of $100\mathrm{m}$is $3800\mathrm{Cycles}$.