(adsbygoogle = window.adsbygoogle || []).push({}); 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.

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# Molar Heat Problem

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