A carnot engine has efficiency of 60%. If the source is at `527^(@)C`, then find the temperature of sink.

Options

A. `700^@C`
B. `350^@C`
C. `1400^@C`
D. no neutral temperature is possible for this thermocouple

Explanation

`E=a theta+b theta^2` (given)
For neutral temperature `theta_n, (dE)/(d theta)=0 implies a+2 theta_a=0`
or `theta_n=-a/(2b) theta_n=-700/2 [ because a/b=700^@C]`
But neutral temperature can never be negative (less than zero) i.e., `theta, 0^@C`
Hence, no neutral temperature is possible for this thermocouple.