# A room contains 48kg of air. How many kilowatt-hoursof energy are necessary to heat the air in the house from 9 degrees Celcius to 27 degrees Celcius? The heat capacity of air is 1.03 Jules per...

A room contains 48kg of air. How many kilowatt-hoursof energy are necessary to heat the air in the house from 9 degrees Celcius to 27 degrees Celcius? The heat capacity of air is 1.03 Jules per gram degrees Celcius.

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### 1 Answer

The total heat necessary to heat up the air in the room from the initial temperature of 9 degree C to the final temperature of 27 degree C is:

`Q =m*c*Delta(T) =48000*1.03*(27-9) =889920 J`

Here the mass was taken in grams since the specific heat of air `c` is given in J/g*K.

The transformation between KWh and J is

`1KW*h =1000*3600 W*s =3.6*10^6 J`

Therefore the total energy in KWh necessary to heat up the air in the room is

`Q (KWh)= (Q(J))/(3.6*10^6) =889920/(3.6*10^6) =0.2472 KWh`

**The total heat necessary to heat up the air in the room is 0.2472 KWh.**

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