Question

A well-sealed room contains 60 kg of air at 200 kPa and 40°C. Now solar energy enters the room at an average rate of 0.8 kJ/s while a 120-W fan is turned on to circulate the air in the room. If heat transfer through the walls is negligible, the air temperature in the room in 30 min will be ____°C.

Answer 1

Answer:

The air temperature in the room in 30 min will be 78.4 °C

Explanation:

The process is done at constant volume (Cv of air: 0.718 kJ/(kg*degC)), so the relation of the heat flow with the temperature of the air is:

$\Delta Q=m*c_v*\Delta T$

The heat flow is the sum of the heat coming from the sun and the heat of the fan:

$\Delta Q = Q_{sun}+Q_{fan}=0.8kJ/s+0.12kJ/s=0.92kJ/s$

In 30 minutes the total amount of heat that enters into the room is

$Q=0.92KJ/s*30min*(60s/min)=1656 kJ$

The variation of temperature in the air can be calculated as

$\Delta T = \frac{Q}{m*c_v}=\frac{1656kJ}{60kg*0.718\frac{kJ}{kg*\circ C} }\\\Delta T =38.4 ^{\circ}  C$

As the initial temperature of the air was 40 deg C, the final temperature will be (40+38.4)=78.4 deg C