Question

A 3 Kg block of Aluminium given 15 kJ of energy and its temperature rises from 30 to 55 Deg Cel. Calculate the value of "c"​

Answer 1

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Answer 2

Mass of the metal, m = 0.20 kg = 200 g

Initial temperature of the metal, T 1= 150 o C

Final temperature of the metal, T 2 = 40 oC

Calorimeter has water equivalent of mass, m = 0.025 kg = 25 g

Volume of water, V = 150 cm 3

Mass (M) of water at temperature T = 27 oC:

150×1=150g

Fall in the-T 2

=150−40=110 o C

Specific heat of water, C w =4.186J/g/K

Specific heat of the metal =C

Heat lost by the metal, =mCT .... (i)

Rise in the temperature of the water and calorimeter system: T 1 −T=40−27=13 oC

Heat gained by the water and calorimeter system: =m 1 C w

T=(M+m)C w

T ....(ii)

Heat lost by the metal = Heat gained by the water and colorimeter system

mCΔT m =(M+m)C w

T w

200×C×110=(150+25)×4.186×13

C=(175×4.186×13)/(110×200)=0.43Jg −1 k −1

If some heat is lost to the surroundings, then the value of C will be smaller than the actual value.