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.