Question

A carnot engine works between ice point and steam point. It is desired to increase efficiency by 20%, by changing temperature of sink to –

• A. 253 K`
• B. 293 K
• C. 303 K
• D. 243 K

Answer 1

Answer: (A)

253 K`

Answer 2

Temp. of source T₁ = 100 + 273 = 373 K.

Temp of sink T₂ = 0 + 273 = 273 K.

Efficiency of carnot engine

h = 1 –$\frac{T_{2}}{T_{1}}$= 1 –$\frac{273}{373}$

h =$\frac{100}{373}$

To increase h¢ =$\frac{100}{373}$+$\frac{100}{373}$×$\frac{1}{5}$=$\frac{120}{373}$

h¢ = 1 –$\frac{T_{2}}{373}$=$\frac{120}{373}$

or T₂ =$\frac{373 - 120}{373}$× 373 = 253 K.

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