Question

a carnot engine has an efficieny of 40%. its efficiency is to be raised to 50%. by how much the temperature of the sink should increase if sink is at 27 deg C.

Answer 1

 η = work W done / Quantity Q of heat absorbed = W / Q
    = 1 - Q2 /Q1
     = 1 - T2 / T1
T1 = source temperature      T2 = sink temperature

η = 40/100 = 1 - T2/T1      =>  T2 = 0.60 T1  = 0.60 * 300K = 180 K
===
New Sink Temperature = T2'

η = 0.50 = 1 - T2'/T1     =>   T2' = 0.50  T1 = 90 K

The temperature of the sink should decrease by  90 deg K or C.