Question

A Carnot engine has efficiency 1/5. Efficiency becomes 1/3 when temperature of sink is decreased by 50 K. What is the temperature of sink?

Answer 1

T1-T2/TO=1/5
4T1=5T2
T2=5/4(T1)
T1-T2-50/TO=1/3
2T1-3T2=150
-0.5T2=150
TO=300K (temp. may never be negative in Kelvin)

Answer 2

Answer:

The temperature of sink is 60 K.

Explanation:

Given that,

Efficiency of engine $\eta=\dfrac{1}{5}$

The formula of efficiency of engine is defined as:

$\eta=1-\dfrac{T_{2}}{T_{1}}$

Where,

$\dfrac{1}{5}=1-\dfrac{T_{2}}{T_{1}}$....(I)

Now, Efficiency becomes 1/3 when temperature of sink is decreased by 50 K.

$\dfrac{1}{3}=1-\dfrac{T_{2}-50}{T_{1}}$...(II)

From equation(I) and (II)

$5T_{2}=4T_{1}$....(III)

$3T_{2}=2T_{1}-150$....(IV)

Now,from equation (III) and (IV)

$2T_{2}=2T_{1}-150$

$2T_{2}-2T_{1}+150=0$

$T_{2}=T_{1}-75$

Put the value of $T_{2}$ in equation (III)

$5\times T_{1}-75=4T_{1}$

$T_{1}=75\ K$

Now put the value of $T_{1}$ in equation (III)

$T_{2}=\dfrac{4\times75}{5}$

$T_{2}=60\ K$

Hence, The temperature of sink is 60 K.