BSc Nursing GNM 2019 Solved Paper

The efficiency of Carnot engine is defined as the ratio of work done to the heat supplied, i.e.
η‌‌=‌

‌ Work done ‌

‌ Heat supplied ‌

‌‌=‌

W

Q1

=‌

Q1−Q2

Q1

‌‌=1−‌

Q2

Q1

=1−‌

T2

T1

Here, T1 is the temperature of source and T2 is the temperature of sink. As given, η=40%=‌

40

100

=0.4 and ‌‌T2=300K So, ‌‌0.4=1−‌

300

T1

⇒‌‌T1=‌

300

1−0.4

=‌

300

0.6

=500K
Let temperatureof the source be increased by x kelvin, then efficiency becomes
η′‌‌=40%+50%‌ of ‌η
‌‌=‌

40

100

+‌

50

100

×0.4
‌‌=0.4+0.5×0.4=0.6
Hence, ‌‌0.6=1−‌

300

500+x

⇒‌‌‌

300

500+x

=0.4
⇒‌‌500+x=‌

300

0.4

=750
x=750−500=250K