Solved Problem on Thermodynamics

The cylinder of a steam engine receives 2300 kcal per unit of time, of this total 2070 kcal are lost to the environment. What is the thermal efficiency of this engine?

Problem data:

Heat received: Q₁ = 2300 kcal;
Heat lost: Q₂ = 2070 kcal.

Problem diagram:

A quantity of heat Q₁ is introduced into the cylinder, doing a work W pushing the piston, a quantity of heat Q₂, unused at work, is expelled as steam (Figure 1).

Figure 1

Solution

The thermal efficiency is given by

\[ \bbox[#99CCFF,10px] {\eta =\frac{W}{Q_{1}}=\frac{Q_{1}-Q_{2}}{Q_{1}}} \]

\[ \begin{gather} \eta =\frac{2300-2070}{2300}\\ \eta=\frac{230}{2300}\\ \eta =0.1=\frac{10}{100} \end{gather} \]

\[ \bbox[#FFCCCC,10px] {\eta =10\;\text{%}} \]

Note: The values are given in kilocalories (kcal), but it is not necessary to consider this factor \( \text{kilo}=1000=10^{3} \), since this value cancels out when dividing the calculation of efficiency.

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