A carnot engine has efficiency 40 % and it absorbs heat from source at `500^@` K. How much should the temperature of source be increased so as to increase its efficiency by 60 % of original efficiency ?

Options

A. 1200 K
B. 750 K
C. 600 K
D. Cannot increases 50% efficiency of carnot engine.

Answer

B

Explanation

`eta_1`=40% =0.4, `T_1`=500 K , `eta_2`=60% = 0.6 , `T._1`=?
`eta_1=1-T_2/T_1`
`T_2/T_1=1-eta_1`
`T_2/500=1-0.4`
`therefore T_2=0.6xx500`
`therefore T_2`=300 K
Now `eta_2=1-T_2/(T._1)`
`T_2/(T._1)=1-eta_2`
`300/(T._1)=1-0.6`
`300/(T._1)=0.4`
`therefore T._1=300/0.4`
`therefore T._1`=750 K