Question 937: Thermodynamics & KTG

Question

The efficiency of an ideal heat engine working
between the freezing point and boiling point
of water, is

Accepted Answer

Efficiency of ideal heat engine,
$\eta  = \left( {1 - {{{T_2}} \over {{T_1}}}} ight)$
Freezing point of water = 0
o
C = 273 K
Boiling point of water = 100
o
C = (100 + 273) K
 = 373 K
$ 	herefore $ Sink temperature, T
2
= 0
o
C = 0 + 273 = 273 K
$ 	herefore $ Source temperature, T
1
= 100
o
C = 100 + 273 = 373 K
Percentage efficiency, %$\eta  = \left( {1 - {{{T_2}} \over {{T_1}}}} ight) 	imes 100$
= $\left( {1 - {{273} \over {373}}} ight) 	imes 100$
= $\left( {{{100} \over {373}}} ight) 	imes 100$ = 26.8%

Answer

26.8%

Answer

20%

Answer

6.25%

Answer

12.5%

Thermodynamics & KTG

The efficiency of an ideal heat engine working between the freezing point and boiling point of water, is

Choose the correct answer: