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A Carnot engine takes 3xx10^6 cal of hea...

A Carnot engine takes `3xx10^6` cal of heat from a reservoir at `627^@C` and gives it to a sink at `27^@C`. The work done by the engine is:

A

`4.2 xx10^6 J`

B

`B.4 xx 10^6 J`

C

`16.B xx10^6 J`

D

zero

Text Solution

Verified by Experts

The correct Answer is:
B
`T_1=627+273=900K`
`Q_1=3xx10^(6) cal`
`T_2=27+273=300K`
Now, from Carnot therom,
`(Q_1)/(T_1)=(Q_2)/(T_2)" or "Q_2=(T_2)/(T_1)xxQ_1 implies Q_2=(300)/(900)xx 3xx10^(6)= 1xx10^(6) cal`
Work done `=Q_1=Q_2`
`=3xx10^(6)-1xx 10^(6) = 2xx10^(6)` cal
`=2xx4.2 xx 10^(6) J = 8.4 xx 10^(6) J`.
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