Collection of Solved Problems in Physics

Efficiency of Carnot Engine

Task number: 1802

• Hint

  Carnot efficiency is given by the relation

  \[\eta=\frac{T_1-T_2}{T_1},\]

  where T₁ is the temperature of hot reservoir and T₂ is the temperature of cold reservoir.

• Analysis

  Write down the relationships for the initial efficiency of the Carnot engine and for the efficiency after changing the temperature of the hot reservoir. Using these equations, evaluate the initial and the final temperature of the hot reservoir. The temperature increase of the hot reservoir can be determined as the difference of these two temperatures.

• Given values

  η1 = 40 % = 0.4	initial efficiency of Carnot engine
  η2 = 50 % = 0.5	efficiency of Carnot engine after increasing the temperature of hot reservoir
  t2 = 9 °C => T2 = 282 K	temperature of cold reservoir
  ΔT1 = ?	temperature increase of hot reservoir

• Solution

  First we write down the relationships for the initial efficiency η₁ of Carnot engine and for the efficiency η₂ after changing the temperature of the hot reservoir:

  \[\eta_1=\frac{T_1-T_2}{T_1}, \qquad \qquad \qquad \eta_2=\frac{T_1^*-T_2}{T_1^*},\]

  where T₁ is the initial temperature of the hot reservoir, T₁^* is the new temperature of the hot reservoir, and T₂ is the temperature of the cold reservoir.

  Now we evaluate the unknown temperatures T₁ and T₁^* from these relations:

  \[T_1=\frac{T_2}{1-\eta_1}, \qquad \qquad \qquad T_1^*=\frac{T_2}{1-\eta_2}.\]

  The unknown temperature difference ΔT₁ of the hot reservoir is given by:

  \[\Delta T_1=T_1^*-T_1\] and by substituting the above evaluated temperatures we obtain \[\Delta T_1=\frac{T_2}{1-\eta_2}-\frac{T_2}{1-\eta_1}.\]

• Numerical solution

  \[\Delta T_1=\frac{T_2}{1-\eta_2}-\frac{T_2}{1-\eta_1}=\left(\frac{282}{1-0.5}-\frac{282}{1-0.4}\right)\,\mathrm{K}\,\dot{=}\, 94\,\mathrm{K}\]

• Answer

  The temperature of the hot reservoir must increase by 94 K, i.e. by 94 °C.