Q.
The temperature of the sink of an engine operating at an efficiency of 0.25 is reduced by 100° C. As a result, the efficiency of heat engine is doubled. Then, the temperature of the source is:
see full answer
Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!
Real Strategies. Real People. Real Success Stories - Just 1 call away
An Intiative by Sri Chaitanya
a
150°C
b
222K
c
242°C
d
400 K
answer is D.
(Unlock A.I Detailed Solution for FREE)
Ready to Test Your Skills?
Check your Performance Today with our Free Mock Test used by Toppers!
Take Free Test
Detailed Solution
Hence, η1=1−T2T2or 0.25=1−T2T2⇒14=1−T2T2T2T2=1−14=34 ….(i)According to question,η2=2η1, and T2'=T2−100∘C∴2×14=1−T2−100T1⇒1−12=T2−100T112=T2T1−100T1⇒34−12=100T1T1=400K
Watch 3-min video & get full concept clarity