Question 278: Thermodynamics & KTG

Question

Work done by a Carnot engine operating between temperatures $$127^{\circ} \mathrm{C}$$ and $$27^{\circ} \mathrm{C}$$ is $$2 \mathrm{~kJ}$$. The amount of heat transferred to the engine by the reservoir is :

Accepted Answer

The efficiency of a Carnot engine is given by:
$$\eta = 1 - \frac{T_{c}}{T_{h}}$$
where:
$$\eta$$ is the efficiency of the Carnot engine,
$$T_{c}$$ is the temperature of the cold reservoir,
$$T_{h}$$ is the temperature of the hot reservoir.
Note that the temperatures must be in Kelvin for this formula.
Here, the given temperatures are $$127^{\circ}C$$ and $$27^{\circ}C$$. Converting these to Kelvin gives:
$$T_{h} = 127^{\circ}C + 273.15 = 400.15 K$$
$$T_{c} = 27^{\circ}C + 273.15 = 300.15 K$$
So the efficiency of the Carnot engine is:
$$\eta = 1 - \frac{300.15}{400.15} = 0.25$$
The efficiency can also be defined as the work done divided by the heat supplied:
$$\eta = \frac{W}{Q_{h}}$$
where:
$$W$$ is the work done by the engine,
$$Q_{h}$$ is the heat transferred to the engine by the hot reservoir.
We can rearrange this equation to solve for $$Q_{h}$$:
$$Q_{h} = \frac{W}{\eta} = \frac{2 \, kJ}{0.25} = 8 \, kJ$$

Answer

8 kJ

Answer

2 kJ

Answer

4 kJ

Answer

2.67 kJ

Thermodynamics & KTG

Work done by a Carnot engine operating between temperatures $127^{\circ} \mathrm{C} and 27^{\circ} \mathrm{C} is 2 \mathrm{~kJ}$. The amount of heat transferred to the engine by the reservoir is :

Choose the correct answer: