Question

2075 Set A Q.No. 106 A Carnot engine has 40% efficiency with a sink at 10°C. By how many degrees should the temperature of the source be increased in order to raise the efficiency to 65%?​

Answer 1

Step-by-step explanation:

efficiency of a carnot engine is given by η = 1 – (T2/T1)

where T1 = source temperature , T2 = sink temperature

T2 = 10⁰C = 273+10 = 282 K

Case I: η = 40%

Therefore, η = 40% = 0.4 = 1 - (283/T1)

So, T1 = 283/0.6 = 471.67 K

Case II: η = 65%

Therefore, η = 65% = 0.65 = 1 - (283/T1)

So, T1 = 283/0.35 = 808.57 K

Difference of temps. of case I and case II = 808.57 - 471.67 = 336.9⁰C

Answer 2

Step-by-step explanation:

Efficiency of a Carnot engine = 1 – (T2/T1) ;

where T1 = source temperature and T2 = sink temperature

T2 = 10⁰C = 273+10 = 283K

The initial efficiency = 40%

So 0.4 = 1 - (283/Initial T1)

Initial T1 = 283/(1-0.4) = 471.67 K

To raise efficiency to 65%

0.65 = 1 - (283/Increased T1)

Increased T1 = 283/(1-0.65) = 808.57 K

The source temperature must be increased by 808.57 - 471.67 = 336.9 K or 336.9°C.