A reversible engine converts one-sixth of the heat input into work. When the temperature of the sink is reduced by `62^@C` , the efficiency of the engine is doubled. The temperatures of the source and sink are

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 _ L ) / (T _ H ) = ( W ) / ( Q _ 1 ) = ( 1 )/( 6 ) `
or ` 5T _ B - 6 T _ L = 0" " `…(i)
` eta _ 2 = 1 - ( T _ L - 62 ) /( T _ H ) = 2 eta _ 1 = ( 1 ) / ( 3 ) ` (Given )
` rArr1 - ( 1 ) /( 3 ) = ( T _ L - 62 ) /( T _ H ) `
or ` 2 T_H - 3 T _ L = - 186 " " `...(ii)
Solving (i) and (ii) we get
` therefore T _ u = 372 K = 99^@ C `
` T _ L = ( 5 ) /( 6 ) T _ H = ( 5 ) /( 6) xx372 K = 310 K = 37^@ C `