# Molar Heat Problem

1. Feb 6, 2007

### Hyari

1. The problem statement, all variables and given/known data
If a heater supplies 1.8 x 10^6 J/h to a room 6.5 m x 4.6 m x 3.0 m containing air at 20 oC and 1.0 atm, by how much will the temperature rise in one hour, assuming no heat losses to the outside?

2. Relevant equations
Q = n*Cv*(Tfinal - Tinitial)
PV = nRT -> n = (PV / RT)
Q = Delta U

3. The attempt at a solution
- It is isochoric. Meaning: volume doesn't change so there is no work involving volume.

(1.8 x 10^6 J / h) (1 calorie / 4.186 J) = 430004.7778 calories

PV = nRT -> n = (PV / RT)

P = 1 atm = 1.013x10^5 N/m^2
V = 6.5 m * 4.6 m * 3.0 m = 89.7 m^3
R = 8.315 (?)
T = 20 + 273.15 = 293.15 K
n ~ 3727.775 mol

---
Q = n * Cv * (Tfinal - Tinitial)

n ~ 3727.775 mol
Cv = 2.98 (cal / mol * K)
T = (Tfinal - 20 oC)
Q = Delta U
Delta U = 430004.7778 calories ?
Q = 430004.7778 calories ?

I'm not sure how to tie this problem together to solve for Tfinal. I'm not sure if I have this correct.

The answer is 22 oC / hour. I cannot get this.

Last edited: Feb 6, 2007