Question

Solve it by using EES

A student living in a 4-m × 6-m × 6-m dormitory room turns on her 150-W fan before she leaves the room on a summer day, hoping that the room will be cooler when she comes
back in the evening. Assuming all the doors and windows are tightly closed and disregarding any heat trans-fer through the walls and the windows, determine the temperature in the
room when she comes back 10 h later. Use specific heat values at room temperature, and assume the room to be at 100 kPa and 15°C in the morning when she
leaves.

Answer 1

Since all the doors and windows are tightly closed.

The system is closed

No change in volume

The fluid is air

Air is considered to be an ideal gas

All the values taken from the ideal gas air table

Initial conditions

T1 = 15 + 273 = 288 K

P1 = 100 kPa

Internal energy U1 = - 7.22 kJ/kg

Volume V = 4 m × 6-m × 6-m = 144 m3

Specific volume v = RT1/P1 = (0.287)(288)/(100)

= 0.8266 m3/kg

Mass = V/v

= 144 m3 / 0.8266 m3/kg

= 174 kg

Energy conservation equation

U2 - U1 = Q + W

For adiabatic case, Q = 0 (disregarding any heat transfer)

U2 = U1 + W*t/m

= - 7.22 kJ/kg + (150*10^-3 kJ/s x 10 hr x 3600s/hr)/(174kg)

= 23.81kJ/kg

Now find the temperature at this internal energy from the table

Temperature T2 = 330.9 K = 57.9 °C