An aluminium kettle weighs 1.05 kg.
(a) what is the heat capacity of the kettle?
(b) how much heat is required to increase the temperature of this kettle from 23.0 to 99.0 degree celcius?
(c) how much heat is required to het this kettle from 23.0 to 99.0 degree celcius if it contains 1.25l of water (densityof 0.997 g/ml and a specific heat of 4.184 j/g degree celcius)

Explained

Explanation:

Given

mass of aluminium kettle()=1.05 kg

(a)Heat capacity of the kettle=Heat capacity of Aluminium

(b)Heat is required to increase the temperature of this kettle from

Q=71.82 kJ

(c)If kettle contains water 1.25 L of water i.e. 1.246 kg of water then heat required to raise the temperature from 23 to 99

Q=71.82+396.20=468.028 kJ

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml