Question

Highest efficiency of heat engine whose lower temperature is 17°C and high temperature of 200°C :

a) 70%
b) 50%
c) 35%
d) 38%

Answer 1

Answer: 38%

Explanation: T1 = 17° C +273= 290K

T2 =200°C +273=473K

ñ= 1-T1/T2×100%

=38%

Answer 2

Answer:

the correct option is (d) 38%

Explanation:

Carnot Cycle : It explains that the ideal reversible thermodynamic cycle where the working substance goes for the successive four phases of isothermal reactions to desired points and adiabatic reactions to desired points.

We know that in a Carnot Cycle,

η = (1 - $\frac{T1}{T2}$ ) x 100%

here it is given that :

T1 = 17°C = 17 + 273k = 290k

T2 = 200°C =200 + 273 = 473K

So putting all these values in the formula, we ghet:

η = (1 - $\frac{290}{473}$) x 100% = 0.3869 x 100% = 38.69 % which can be rewritten as 38%

So, the correct option is (d) 38%

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