A system consists of 3.0 kg of water at 80øC. 25 J of work is done on the system by stirring with a paddle..?

Question

wheel, while 63 J of heat is removed. What is the change in internal energy of the system?

Answer 1

Use the first law of thermodynamics, which states that the change in internal energy of a system is equal to the work done *on* the system by the environment plus the heat added to the system:

delta-U = w + delta-Q

(One will sometimes see this law written with the "work" term defined as the work done *by* the system on the environment, in which case the sign in front of the work term is negative: delta-U = -w + delta-Q)

Using the first law as I first wrote it, we have that w = +25 J, and delta-Q = -63 J (negative because heat is removed, not added, to the system)

delta-U = 25 J - 63 J = -38 J

Answer 2

The equation you need is

ΔQ = m c ΔT

ΔQ = heat added or removed
m = mass
c = specific heat
ΔT = change in temperature

Your problem says that ΔQ = +25J - 63J = -38J

You don't ask, but you can now compute the temperature change of the water using the equation above.