A carnot engine converts one sixth of the heat input into work. The sink temperature is reduced by `62^@ C`, the efficiency of engine gets doubled. Find the source and the sink temperature.

Options

A. `99^@C,37^@C`
B. `80^@C, 37^@C`
C. `95^@C, 37^@C`
D. `90^@C, 37^@C`

Answer

A

Explanation

`eta_1=1-T_2/T_1` and `eta_1=W/Q_1=1/6`
`1/6=1-T_2/T_1`
`therefore T_1=6T_1-6T_2`
`therefore 5T_1-6T_2=0` ….(1)
Now `eta_2=1-T_2/T_1`
`2xx1/6=1-(T_2-62)/T_1`
`1/3=1-(T_2-62)/T_1`
`T_1=3T_1-3T_2+186`
`therefore 2T_1-3T_2=-186`....(2)
From equation (1) and (2),
`2xx(6T_2)/5-3T_2=-186`
`therefore 12T_2-15T_2=-930`
`therefore -3T_2=-930`
`therefore T_2`=310 K
`rArr` From equation (1) ,
`T_1=(6T_2)/5 =(6xx310)/5`=372 K
`therefore t_2=T_2-273` and `t_1=T_1-273`
`therefore t_2=310-273` and `t_1=372-273`
`therefore t_2=37^@C` and `t_1=99^@C`