Question

In a certain steam engine, the boiler temperature is \(127^{\circ} \mathrm{C}\) and the cold reservoir temperature is \(27^{\circ} \mathrm{C} .\) While this engine does \(8.34 \mathrm{kJ}\) of work, what minimum amount of heat must be discharged into the cold reservoir?

Short answer

Answer: The minimum amount of heat that must be discharged into the cold reservoir is \(25.02\,\mathrm{kJ}\).

Step-by-step solution

Converting temperatures to Kelvin

We need to convert the given temperatures from Celsius to Kelvin, as we will be working with the absolute temperature in our calculations. To do this, we add 273.15 to the Celsius temperature: $$ T_{h} = 127^{\circ} \mathrm{C} + 273.15 = 400.15\,\mathrm{K} $$ $$ T_{c} = 27^{\circ} \mathrm{C} + 273.15 = 300.15\,\mathrm{K} $$

Calculating Carnot efficiency

Now we will calculate the Carnot efficiency using the formula \(1-\frac{T_{c}}{T_{h}}\): $$ \eta_{Carnot} = 1-\frac{T_{c}}{T_{h}} = 1-\frac{300.15}{400.15} = 1-0.75 = 0.25 $$

Finding the heat absorbed from the hot reservoir

Next, we will find the heat absorbed from the hot reservoir using the relationship between work done (W), heat absorbed (Q) and efficiency: $$ \eta_{Carnot} = \frac{W}{Q} \implies Q = \frac{W}{\eta_{Carnot}} $$ Plugging in the values, $$ Q = \frac{8.34\,\mathrm{kJ}}{0.25} = 33.36\,\mathrm{kJ} $$

Calculating the heat discharged in the cold reservoir

Finally, we can find the heat discharged in the cold reservoir using the conservation of energy principle. The heat absorbed from the hot reservoir minus the work done equals the heat discharged to the cold reservoir: $$ Q_{c} = Q - W $$ Plugging in the values, $$ Q_{c} = 33.36\,\mathrm{kJ} - 8.34\,\mathrm{kJ} = 25.02\,\mathrm{kJ} $$ Hence, the minimum amount of heat that must be discharged into the cold reservoir is \(25.02\,\mathrm{kJ}\).