Question

4,020 J of heat is added to a system while 4,922 J of work is done on the system. What is the change in internal energy?

Answer 1

From `1` st law of thermodynamics, heat energy added `dQ=4020 J`, and work done on the system `-dW=4922J`,so change in internal energy `dU`, is related as,

`dQ = dU + (-dW)`

So, `dU = dQ+dW = 4020+4922=8942J`

So,internal energy has increased by `8942 J`