1.

The efficiency of a carnot engine working between temperature T_(1) and T_(2) is eta. It will be also eta if it works between temperatures

Answer»

`T_(1) + 10` and `T_(2) + 10`
`T_(1) - 10` and `T_(2) - 10`
`2T_(1)` and `2T_(2)`
in all the above CASES

Solution :`ETA = 1 - (T_(2))/(T_(1))`
If `T_(1). = 2T_(1)` and `T._(2) = 2T_(2)`, then `eta. = 1-(T._(2))/(T._(1)) = 1 - (2T_(2))/(2T_(1)) = eta`
In other cases, it is not so.


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