KCET 2018 Physics Solved Paper

Given, heat taken from source, Q1=300 calories; temperature of source T1=500K; heat rejected to the sink,Q2=150 calories.
Temperature of sink, T2=?
Now, efficiency, η=‌

Q1−Q2

Q

Also, efficiency, η=‌

T1−T2

T1

Therefore,‌

Q1−Q2

Q1

=‌

T1−T2

T1

⇒‌

300−150

300

=‌

500−T2

500

⇒‌

150

300

=1−‌

T2

500

⇒‌

1

2

=1−‌

T2

500

⇒‌

T2

500

=1−‌

1

2

⇒T2=500×‌

1

2

=250K
Thus, temperature of sink = 250 K