In a certain process, 400 cal of heat are supplied to a system and the same time 105 J of mechanical work done on the system . The increase in its internal energy is

Options

A. 20 cal
B. 303 cal
C. 404 cal
D. 425 cal

Answer

D

Explanation

Here, `dQ=400"cal",dW=-105 J`
= `-105//4.2"cal"=-25"cal",dU` =?
`"Now"" " du=dQ-dW=400-(-25)=425"cal"`
Note dW is negative because ,work done on the system .