Question 736: Thermodynamics & KTG

Question

Heat engine operating between temperature $T_1$ and $T_2$ has efficiency $\frac{1}{6}$. When $T_2$ is lowered by 62 K , its efficiency increases to $\frac{1}{3}$. Then $T_1$ and $T_2$ respectively are

Accepted Answer

Given:
The initial efficiency of the heat engine is $\eta_1 = \frac{1}{6}$, which can be expressed as:
$ \eta_1 = 1 - \frac{T_2}{T_1} \quad \Rightarrow \quad T_2 = \frac{5}{6} T_1 \quad 	ext{...(i)} $
When the lower temperature $T_2$ is decreased by 62 K, the efficiency increases to $\eta_2 = \frac{1}{3}$. This can be expressed as:
$ \eta_2 = \frac{1}{3} = 1 - \frac{T_2 - 62}{T_1} $
Substituting from equation (i):
$ \frac{\left(\frac{5}{6} T_1 - 62ight)}{T_1} = 1 - \frac{1}{3} $
Simplifying this, we get:
$ \frac{5}{6} T_1 - 62 = \frac{2}{3} T_1 $
Rearranging terms, we find:
$ \frac{5}{6} T_1 - \frac{2}{3} T_1 = 62 $
Solving this equation will give:
$ \frac{1}{6} T_1 = 62 $
Thus:
$ T_1 = 372 \, 	ext{K} $
Substituting back into equation (i) to find $T_2$:
$ T_2 = \frac{5}{6} 	imes 372 = 310 \, 	ext{K} $
Thus, the temperatures $T_1$ and $T_2$ are 372 K and 310 K, respectively.

Answer

$372 \mathrm{~K}, 310 \mathrm{~K}$

Answer

$372 \mathrm{~K}, 330 \mathrm{~K}$

Answer

$330 \mathrm{~K}, 268 \mathrm{~K}$

Answer

$310 \mathrm{~K}, 248 \mathrm{~K}$

Thermodynamics & KTG

Heat engine operating between temperature T_1 and T_2 has efficiency \frac{1}{6}. When T_2 is lowered by 62 K , its efficiency increases to \frac{1}{3}. Then T_1 and T_2 respectively are

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