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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. |
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Answer» `99^@C,37^@C` `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` |
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