NEET Class neet Answered
An engine has efficiency 0.25 when temperature of sink is reduced by 58℃, if it's efficiency is doubled , then the temperature of source is :-
Asked by jhajuhi19 | 26 Jun, 2019, 09:01: AM
Expert Answer
I assume the question is, initially the engine works with efficiency 0.25 and efficeiency is doubled
after reducing the sink temperature by 52°C. It is required to get the source temperature.
Initial effieciency ηi of engine working with source temperature T1 and sink temperature T2 is given by,
ηi = 1 - ( T2 / T1 ) = 0.25 ..............(1)
changed efficiency ηf after reducing the sink temperature,
ηf = [ 1 - (T2 - 58) / T1] = 0.5 ........................(2)
eqn.(2) is rewritten as, 1 - ( T2 / T1 ) + (58 / T1) = 0.5 ...............(3)
using eqn.(1) and eqn.(3), we get 58/T1 = 0.25 or source temperature T1 = 232 K
Answered by Thiyagarajan K | 26 Jun, 2019, 11:34: AM
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