Calculate the number of kJ of heat neccesary to raise the temperature of 60.0g of aluminum from `35^(@) C` to `55^(@)C`. Molar heat capacity of Al is 24 J `mol^(-1) k^(-1)`.

Explanation

We are given that
Molar heat capacity of Al = 24 `mol^(-1)K^(-1)`
Mass = 60.0 g
Temperature changes = `Delta T = (273 + 35 - 273 + 55)`
= 308 - 328 = 20 K
We are to calculate the heat required in k.J
We know that
Amount of heat required to raise the temperature 1 mole of substance by `1^(@)` degree is the molar heat capacity.
i.e., 27 g or Al requires heat = 24
60 g of Al requires heat = `(24)/(27) xx 60 = (480)/(9)`
Amount of heat required q = `C Delta T`
So, `q = (480)/(9) xx 20 = (9600)/(9) = 1066.4 j`
Amount of heat required in Kj = 1.07 kj