Question 696: Thermodynamics & KTG

Question

An engine operating between temperatures $T_1$ and $T_2$ has efficiency $\frac{1}{5}$. When $T_2$ is lowered by 45 K , its efficiency becomes $\frac{1}{2}$. Temperatures $T_1$ and $T_2$ are respectively

Accepted Answer

Let $T_1$ be the higher temperature and $T_2$ be the lower temperature.
For a Carnot engine, efficiency $\eta$ is given by:
$ \eta = 1 - \frac{T_2}{T_1} $
Step 1: Using first condition ($\eta = \frac{1}{5}$)
$ 1 - \frac{T_2}{T_1} = \frac{1}{5} $
$ \frac{T_2}{T_1} = 1 - \frac{1}{5} = \frac{4}{5} $
$ T_2 = \frac{4}{5} T_1 $
Step 2: Second condition ($T_2$ decreased by $45~K$, efficiency $\frac{1}{2}$)
New lower temperature $T_2' = T_2 - 45$.
$ \eta' = 1 - \frac{T_2 - 45}{T_1} = \frac{1}{2} $
$ 1 - \frac{T_2 - 45}{T_1} = \frac{1}{2} $
$ \frac{T_2 - 45}{T_1} = 1 - \frac{1}{2} = \frac{1}{2} $
$ T_2 - 45 = \frac{1}{2}T_1 $
Step 3: Solving the two equations
From above:
$T_2 = \frac{4}{5}T_1$
$T_2 - 45 = \frac{1}{2}T_1$
Substituting $T_2$ from first into second:
$ \frac{4}{5}T_1 - 45 = \frac{1}{2}T_1 $
$ \frac{4}{5}T_1 - \frac{1}{2}T_1 = 45 $
Take common denominator:
$ \frac{8 - 5}{10}T_1 = 45 $
$ \frac{3}{10}T_1 = 45 $
$ T_1 = 45 	imes \frac{10}{3} = 150~K $
$ T_2 = \frac{4}{5} 	imes 150 = 120~K $
Step 4: Final answer
$ \boxed{T_1 = 150~K, \quad T_2 = 120~K} $
Answer: Option D

Answer

$100 \mathrm{~K}, 70 \mathrm{~K}$

Answer

$160 \mathrm{~K}, 120 \mathrm{~K}$

Answer

$140 \mathrm{~K}, 110 \mathrm{~K}$

Answer

$150 \mathrm{~K}, 120 \mathrm{~K}$

Thermodynamics & KTG

An engine operating between temperatures T_1 and T_2 has efficiency \frac{1}{5}. When T_2 is lowered by 45 K , its efficiency becomes \frac{1}{2}. Temperatures T_1 and T_2 are respectively

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