Question

2. A student living in a 3-m × 4-m × 4-m dormitory room turns on her 100-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 8 h later. Use specific heat values at room temperature and assume the room to be at 100 kPa and 20°C in the morning when she leaves.

Answer 1

The room temperature will be
`90.3^(0)`

Step-by-step explanation:

We know from First law of thermodynamics that the amount of heat flow (Q) through a system is given by

`Q = m C_(V)\Delta T`

where, 'm' is the mass of the system,
`C_(V)` is the specific heat and
`\Delta T` is the temperature difference. Also we know,
`C_(V)` = 0.718 KJ
`Kg^(-1)`.

According to the problem, the heat flow can also be written as

`Q = P * t = 100 W * 8 hr = 0.1 kW * 8 * 3600 sec = 2880 KJ`

So,

`&& mC_(V)\Delta T = 280000\\&or,& \Delta T = (280000)/(m * C_(V))`

Again, if 'm' be the mass of the gas and 'R' be the Universal gas constant =
`0.287 KPam^(3)Kg^{-1K^(1)}`then from Ideal gas equation we can write,

`&& PV = mRT\\\\&or,& m = (PV)/(RT) = (100kPa * (3 * 4 * 4))/(0.287 KPam^(3)Kg^(-1)K^(-1) * 293) = (100 * 48)/(0.287 * 293) = 57.08 Kg`

So,

`&& \Delta T = (2880)/(57.08 * 0.718)\\\\&or,& T_(2) - T_(1) = 70.3\\\\&or,& T_(2) = T_(1) + 70.3 = 293 + 70.3 K = 363.3 K\\\\&or,& T_(2) = 363.3 - 273 = 90.3^(0) C`