Thermal Dynamics Problem

[interXdragon]

1. The problem statement, all variables and given/known data

Q = ?
ΔU = + 8.0x10^3
W = perhaps 0 since isovolumetric?
Energy Lost = -2.0x10^3
Mass of Water = 2.0 kg


2. Relevant equations

I'm stuck on this thermal dynamics problem. It states: "A 2.0 kg quantity of water is held at constant volume in a puressure cooker and is heated by a range elment. The system's interal energy increases by 8.0 x 10^3 J. However, the pressure cooker is not well insulated, and 2.0 x 10^3 J of energy is transferred to the surrounding air. How much energy is trasferred from te range elment to the pressure cooker as heat?"


3. The attempt at a solution

I know that is is an isovolumetric problem, and in my book, that means work is equal to zero. Since 'w' is taken out of the equation, I solved it by Q = 8.0x10^3 J, from ΔU = Q - W. I then subtracted that by 2.0x10^3 J. My answer was completely off. What am I thinking wrong?

[Hootenanny]

Your on the right track. Consider the first law of thermodynamics, which are you correctly state is \Delta U = Q - W and as you correctly say, the process is isochoric \Rightarrow W = 0. So now we have \Delta U = Q where Q is the heat added to the system; this is the heat transfered to the pressure cooker which raised to internal energy. Now, if Q' is the heat transfered from the element is the pressure cooker and the surrounding air, then Q'=Q+2.0x103. Does that make sense?

[interXdragon]

thanks to that, now i have the right answer! but I'm still confused on the last sentence. i don't see why we should include 2.0 x 10^3 J into the transferred heat.

[Hootenanny]

thanks to that, now i have the right answer! but I'm still confused on the last sentence. i don't see why we should include 2.0 x 10^3 J into the transferred heat.
Okay, you are given that the internal energy of the pressure cooker is increased by \Delta U, from the first law of thermodynamics you can calculate that Q joules of heat must be transferred to the pressure cooker from the element (in this case [isochoric] note that \Delta U = Q). Now, the element also supplies 2.0 x 10^3 J to the air so the total heat the element supplies is the sum of those two values.