Chemistry - Thermodynamics - II Question with Solution | TestHub

Question Explanation: Calculate the temperature of source and sink using Carnot engine efficiency data before and after changing sink temperature.

Concept: Carnot engine efficiency $\left(\eta =1-\frac{{\mathrm{T}}_{\mathrm{L}}}{{\mathrm{T}}_{\mathrm{H}}}\right)$

Solution:

$$\begin{gathered} \text{displaylines}\left(\eta =1-\frac{{\mathrm{T}}_{\mathrm{L}}}{{\mathrm{T}}_{\mathrm{H}}}\right) \\ \frac{1}{6}=1-\frac{{\mathrm{T}}_{\mathrm{L}}}{{\mathrm{T}}_{\mathrm{H}}}\dots \dots \left(1\right) \\ \frac{1}{3}=1-\frac{\left({\mathrm{T}}_{\mathrm{L}}-65\right)}{{\mathrm{T}}_{\mathrm{H}}}\dots \left(2\right) \\ \frac{1}{3}=\frac{1}{6}+\frac{65}{T_H} \end{gathered}$$

T_H = 6 × 65 = 390 K = 117°C

${\mathrm{T}}_L=\frac{5}{6}\times 390=325K=52$°C