The heat capacity of air is much smaller than that of water,and relatively modes

Question

The heat capacity of air is much smaller than that of water,and relatively modest amounts of heat are needed to change itstemperature. This is one of the reasons why desert regions,though very hot during the day, are bitterly cold atnight. The heat capacity of air at room temperature andpressure is approximatley 21 J K-1mol-1. How much energy is required to raise thetemperature of a room of dimensions 5.5 m 6.5 m 3.0 mby 10?? If losses are neglected, how long will it take aheater rated at 1.5 kW to achieve that increase given that?

my solutions manual says to first calculate for moles of air and ituses pV=nRT. For the temperature value it uses 283 K. Why does it use this temperature when it said it was at room temp(298 K)? The heat capacity of air is much smaller than that of water,and relatively modest amounts of heat are needed to change itstemperature. This is one of the reasons why desert regions,though very hot during the day, are bitterly cold atnight. The heat capacity of air at room temperature andpressure is approximatley 21 J K-1mol-1. How much energy is required to raise thetemperature of a room of dimensions 5.5 m 6.5 m 3.0 mby 10?? If losses are neglected, how long will it take aheater rated at 1.5 kW to achieve that increase given that?

my solutions manual says to first calculate for moles of air and ituses pV=nRT. For the temperature value it uses 283 K. Why does it use this temperature when it said it was at room temp(298 K)?

Explanation / Answer

Room temperature is not a defined temperature inscience. So the best thing to do is to specify in yourcalculateing what the starting temperature was. I willuse the 283 that the books says even though you can use anyreasonable temperature. Room pressure will always be 1 atm. The volume you needto calculate first. 5.5*6.5*3.0=107.25 m^3 Then convert that to cm^3 107.25*100^3=107250000 cm ^3. One cm^3 is equal to on mL. 107250000 mL/1000 mL/L=107250 L are in the room. So the number of moles in the room is 1 atm *107250 L = n * .0821 * 283 K n=4616 moles. Now to find the number of Joules needed just time the numberof moles by the heat capacity and delta T which in this case is10. 21*10*4616=969360 J = 969.360 kJ Now for the second part watt is just J/s. Dividethe number of kJ needed by the amount of kW provided to get thenumber of seconds it would take. 969.360/1.5=646.24 s Which is about 10 min and 46 seconds. Which is your final answer.

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