Question

The temperature of sink is 280 k for heat engine and its efficiency is 50 % what will be the temperature of source

Answer 1

temperature of sink (T2) = 280k
efficiency= 50%
temperature of source is T1 is ?

efficiency= (T1-T2)/T1
efficiency=1-T2/T1
T2/T1= 1- efficiency
T1 = T2/(1- efficiency)
T1 = 280/(1-0.5)
T1 = 280/0.5
T1 = 560 k

Answer 2

Answer:

The temperature of source will be 560K for efficiency of engine50%.

Explanation:

Consider that

The temperature of source = $T_{H}$

The temperature of sink =  $T_{c}$

The efficiency of engine =50% = 0.5

We know that    

Efficiency of engine   $=\frac{T_{H}-T_{c}}{T_{H}}$

                        $0.5=1-\frac{280K}{T_{H}} \\ \\ 1-0.5=\frac{280K}{T_{H}}\\ \\ T_{H}=\frac{280K}{0.5}\\ \\ T_{H}=560K$

Therefore the temperature of source equals to 560K.