Solution Found!
A computer in a closed room of volume 5000 ft3 dissipates
Chapter 3, Problem 293EUP(choose chapter or problem)
Problem 293EUP
A computer in a closed room of volume 5000 ft3 dissipates energy at a rate of 10kw. The room has 100 lbm of wood, 50 lbm of steel, and air, with all material at 540 R, 1 atm. Assuming all of the mass heats up uniformly, how much time will it take to increase the temperature by 20 F?
Questions & Answers
QUESTION:
Problem 293EUP
A computer in a closed room of volume 5000 ft3 dissipates energy at a rate of 10kw. The room has 100 lbm of wood, 50 lbm of steel, and air, with all material at 540 R, 1 atm. Assuming all of the mass heats up uniformly, how much time will it take to increase the temperature by 20 F?
ANSWER:
Solution 293EUP
Step 1 of 3:
The temperature inside a closed room is increased by 20 F as the computer releases heat. We are going to find the time taken for this process.
The volume of the room V = 5000 ft3 = 142 m3
The mass of wood m1 = 100 lbm = 45.4 kg
The mass of steel and air m2 = 50 lbm = 22.7
The initial temperature T1 = 540 R = 80 F = 300 K
The final temperature T2 = 80 F + 20 F = 100 F = 311 K
The change in the temperature ΔT = 11 K
The heat dissipated = 10 kW
From the table A.3,
The density of wood ⍴1 = 510 m3
The density of steel ⍴2 = 7820 m3
The specific heat of wood C1 = 1.38 kJ/kg.K
The specific heat of steel C2 = 0.46 kJ/kg.K
From the table A.5,
The specific heat of air Cv = 1.38 kJ/kg.K
The density of steel ⍴air = 1.169 m3