An ideal gas heat engine operates in Carnot cycle between `227^(@)C` and `127^(@)C`. It absorbs `6 x 10^(4) cals` of heat at higher temperature. Amount of heat converted to work is

Options

A. `2.4 xx 10^(4) cal`
B. `6 xx 10^(4) cal`
C. `1.2 xx 10^(4) cal`
D. `4.8 xx 10^(4) cal`

Answer

C

Explanation

We have `Q/T` = consta nt `implies (Q_2)/(Q_1)=(T_2)/(T_1)`
Given, `Q_(1)=6xx10^(4) cal`,
`T_(1)=227+273=500K`
`T_(2) = 127+273 = 400K`
`:. (Q_2)/(6xx10^4)= 400/500`
`implies Q_(2) = 4/5xx6xx10^(4)=4.8xx10^(4)cal`
Now heat converted to work
`=Q_(1) - Q_(2) = 6.0xx10^(4)-4.8xx10^(4)=1.2xx10^(4)cal`
Aliter: `eta = (T_1-T_2)/(T_1) = W/Q implies W = (Q(T_1-T_2))/(T_1)`
`W = (6xx10^(4)[(227+273)-(273+127)])/((227+273))`
`implies W = (6xx10^(4)xx100)/(500)=1.2xx10^(4)cal`.