An ideal gas is taken through the process `ABC` as shown in figure. If the internal energy of the substance decreases by `10000J` and a heat of `7159` cal is released by the system, calculate the value of meachanical equivalent of heat `(J)`.

Answer

`(30000)/(7159) = 4.19 J//cal`

Explanation

`DeltaW_(ABC) = DeltaW_(AB) +DeltaW_(BC)`
`DeltaW_(ABC) = 0 +500 xx 10^(3) xx 0.04 = 20000J`
`DeltaQ_(ABC) = DeltaW_(ABC) +DeltaU_(ABC)`
`DeltaQ_(ABC) = 20000 +10000 = 30000J`
`J = (DeltaQ(J))/(Delta Q(cal)) = (30000J)/(7159cal) = 4.19J//cal`.