Someone leaves a 100-W light bulb on before leaving and sealing up a 2 m x 4 m x 4 m room full of air. Initially the air temperature is 20oC, and the air pressure is 100 kPa. Assuming that the mass of the air in the room stays constant and ignoring any pressure changes

Question

(a) determine the temperature of the air in the room after 8 hours.
(b) What would the temperature of the room be after 8 hours if someone had also left a 50-W fan operating in the room at the same time?
Given: Air; L = 4m; W = 4m; H = 2m; T1 = 20oC = 293 K; W?=?0.1 kW; P = 100 kPa; t = 8 hours
Assume: Q = 0 (insulated); ?KE = ?PE = 0
Air behaves as an ideal gas with constant specific heats: cv = 0.718 kJ/kg-K
R = 0.287 kJ/kg-K
P = constant; m = constant
What will be an ideal response?

Answer 1

Solution: This is a closed system. Therefore, beginning with the First Law for closed systems, the First Law can be reduced to
W = m (u1 – u2)
(a)
In 8 hours, the total work is W = W?t = (-0.1 kW) (8 hr) (3600 s/hr) = -2880 kJ
V = LWH = 32 m3
m = PV/RT1 = 38.05 kg
W = m cv(T1 – T2)
Solving: T2 = 125oC
(b) With the fan running, the power input is now W? = -0.1 kW + -0.05 kW = -0.15 kW
Then, W = -4320 kJ
And solving for T2 = 178oC

Fortunately for many people, rooms are not perfectly insulated, and heat losses would occur to keep the temperature lower. In addition, any other mass inside a room would be absorbing heat and slowing the temperature increase.