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

Answer:

The room temperature will be $90.3^{0}$

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

$Q = P \times t = 100 W \times 8 hr = 0.1 kW \times 8 \times 3600 sec = 2880 KJ$

So,

$&& mC_{V}\Delta T = 280000\\&or,& \Delta T = \frac{280000}{m \times C_{V}}$

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,

$&& PV = mRT\\\\&or,& m = \frac{PV}{RT} = \frac{100kPa \times (3 \times 4 \times 4)}{0.287 KPam^{3}Kg^{-1}K^{-1} \times 293} = \frac{100 \times 48}{0.287 \times 293} = 57.08 Kg$

So,

$&& \Delta T = \frac{2880}{57.08 \times 0.718}\\\\&or,& T_{2} - T_{1} = 70.3\\\\&or,& T_{2} = T_{1} + 70.3 = 293 + 70.3 K = 363.3 K\\\\&or,& T_{2} = 363.3 - 273 = 90.3^{0} C$