Question

A carnot engine whose low temperature resevoir is 200k has an efficiency of 50%.it is desired to increase this to 75% by how many degree must the temperature reservior be decreased if the temperature of the heigher temperature reservoir remains constant?

Answer 1

Efficiency of a Carnot cycle is given by :

η =  Wnet/Qinput =  (Q1−Q2)/Q1=1−Q2/Q1  

Using Clausius Inequality (Since Carnot cycle is reversible cycle),

η =  1−T2/T1  

Note : Put the temperature values in absolute scale (Kelvin scale)

Case 1 : Given Efficiency = 50%, T1 = 7°C=280 K,

Using the above formula we get the higher temperature T2 = 560 K

Case 2 : Given Efficiency = 70% , T1 = 7°C= 280K

Using the above formula we get higher temperature T2 = 933.33 K

Therefore the temperature of high temperature reservoir should be increased by = 933.33 - 560 = 373.33 K or 373.33 °C

Answer 2

Explanation:

ansans- 373.33k thank you for your self