A substance is taken through the process abc as shown in fig, if the internal energy of the substance increase by 5000 J and a heat of 2625 cal is given to the system, calculate the value of J.

Answer

A, C, D

Explanation

Now, `DeltaQ = 2625cal .`
` DeltaU = 5000J
Form graph,`
`Delta W = 200xx10^3xx0.03`
` = 6000J`
`Now Delta Q = DeltaW+DeltaU`
`rArr 2625cal = 6000+5000J`
`rArr J = (11000)/(2625) = 4.19J/cal.