The efficiency of Car not engine is 50 % and temperature of sin k is 500 K. If the temperature of source is kept constant and its efficiency is to be raised to 60 %, then the required temperature of the sink will be:

Question

The efficiency of Car not engine is 50 % and temperature of sin k is 500 K. If the temperature of source is kept constant and its efficiency is to be raised to 60 %, then the required temperature of the sink will be: (A) 600 K (B) 400 K (C) 500 K (D) 100 K

Tardigrade Admin · Accepted Answer

Efficiency of the Carnot engine is given by

η = 1 - \frac{T _{2}}{T _{1}} ..

(i)
where

T_{1} =

temperature of source

T_{2} =

temperature of

sin k

Given,

η = 50

Substituting in relation (i), we have or

0.5 = 1 - \frac{500}{T _{1}}

or

\frac{500}{T _{1}} = 0.5

∴ T_{1} = \frac{500}{0.5} = 1000 K

Now, the temperature of sink is changed to

T_{2}

and the efficiency becomes

60%

i.e,

0.6

. Using relation (i), we get

0.6 = 1 - \frac{T _{2}}{1000}

or

\frac{T _{2}}{1000} = 1 - 0.6 = 0.4

or

T_{2} = 0.4 \times 100 = 400 K

Carnot engine is not a practical engine because many ideal situations have been assumed while designing this engine which can practically not be obtained.

Q. The efficiency of Car not engine is 50% and temperature of sink is 500K. If the temperature of source is kept constant and its efficiency is to be raised to 60%, then the required temperature of the sink will be:

600 K

400 K

500 K

100 K

Q. The efficiency of Car not engine is $50%$ and temperature of $sin k$ is $500 K$ . If the temperature of source is kept constant and its efficiency is to be raised to $60%$ , then the required temperature of the sink will be: