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Answer :
Final answer:
To reduce the room temperature from 315 K to 295 K, the energy that needs to be transferred out of the room is approximately 25377600 Joules, calculated using the specific heat formula Q = mcΔT.
Explanation:
To find the energy that must be transferred out of the room in order to decrease the temperature of the air from 315 K to 295 K, we need to use the specific heat formula: Q = mcΔT. The formula represents the amount of heat (Q) that is exchanged by a body in modifying its temperature, where m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
Here, the room volume is 7 m x 7 m x 3 m = 147 m3. Assuming the density of air to be 1.2 kg/m3, the mass of the air inside the room is therefore 1.2 kg/m3 x 147 m3 = 176.4 kg. The specific heat capacity at constant volume c is given as 720 J/kg-K, and the decrease in temperature ΔT is 315 K - 295 K = 20 K.
By substituting these values into the formula, we get Q = 176.4 kg x 720 J/kg-K x 20 K = 25377600 J. Therefore, an amount of approximately 25377600 Joules needs to be transferred out of the room to achieve the desired decrease in temperature.
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