Question

A Carnot engine whose low temperature reservoir is at 7°C has an efficiency of 50%. It is desired to increase the efficiency to 70%. By how many degrees should the temperature of the high temperature reservoir be increased?

Options

• 840 K

• 280 K

• 560 K

• 380 K

Answer 1

380 K

Explanation:

T₂ = 7°C = (7 + 273) = 280 K

η = `1-"T"_2/"T"_1` 

⇒ `"T"_2/"T"_1` = 1 - η

= `1-50/100=50/100=1/2`

∴ T₁ = 2 × T₂ = 2 × 280 = 560 K

New efficiency, η' = 70%

∴ `"T"_2/"T"_1` = 1 - η' = `1 - 70/100=30/100=3/10`

∴ `"T"_1=10/3xx280=2800/3`

= 933.3 K

∴ Increase in the temperature of high temp. reservoir

= 933.3 - 560

= 373.3 K = 380 K