Solveeit Logo
Login

Question

Question: The efficiency of the Carnot engine is \(\dfrac{1}{5}\) . Now the temperature of the source is incre...

The efficiency of the Carnot engine is 15\dfrac{1}{5} . Now the temperature of the source is increased by 100∘ C{100^ \circ }\,C then efficiency becomes doubled. The initial temperature of sink and source are respectively.
(1) 240 K, 360 K240\,K,\,360\,K
(2) 240 K, 300 K240\,K,\,300\,K
(3) 200 K, 250 K200\,K,\,250\,K
(4) 300 K, 240 K300\,K,\,240\,K

Explanation

Solution

Use the below formula of the efficiency, and substitute the parameters and the conditions given in the questions. Based on the two conditions, form two equations from and solve the two equations to find the value of the sink and the source temperature.

Useful formula:
The efficiency of the Carnot’s engine is given by

η=T1−T2T1\eta = \dfrac{{{T_1} - {T_2}}}{{{T_1}}}

Where η\eta is the efficiency of the Carnot’s engine, T1{T_1} is the temperature of the sink and T2{T_2} is the temperature of the source.

Complete step by step solution:
It is given that the efficiency of the Carnot’s engine, η=15\eta = \dfrac{1}{5}
The temperature of the source increased by 100∘ C{100^ \circ }\,C , then the efficiency became doubled.
Using the formula of the efficiency,

η=T1−T2T1\eta = \dfrac{{{T_1} - {T_2}}}{{{T_1}}}

Substituting the initial conditions,

15=T1−T2T1\dfrac{1}{5} = \dfrac{{{T_1} - {T_2}}}{{{T_1}}} --------------(1)

Substituting the final condition in the formula, we get

25=(T1+100)−T2(T1+100)\dfrac{2}{5} = \dfrac{{\left( {{T_1} + 100} \right) - {T_2}}}{{\left( {{T_1} + 100} \right)}} ------------(2)
Solving equation (1) and (2), we get

T1−T2T1=(T1+100)−T22(T1+100)\dfrac{{{T_1} - {T_2}}}{{{T_1}}} = \dfrac{{\left( {{T_1} + 100} \right) - {T_2}}}{{2\left( {{T_1} + 100} \right)}}

By simplifying the above equation, we get

2(T1+100)(T1−T)2=T1(T1+100)−T22\left( {{T_1} + 100} \right){\left( {{T_1} - T} \right)_2} = {T_1}\left( {{T_1} + 100} \right) - {T_2}
T1=240 K{T_1} = 240\,K

Substituting the value of the source temperature in the equation (1), we get

15=240−T2240\dfrac{1}{5} = \dfrac{{240 - {T_2}}}{{240}}

By simplifying the above equation, we get

T2=300 K{T_2} = 300\,K

Hence the temperature of the sink is obtained as 300 K300\,K.

Thus the option (2) is correct.

Note: Normally the sink temperature represents the cold environment and the source represents the hot environment. The practical examples of this are the heat engine is the source and the natural environment represents the sink that cools the heat.