Question

a sphere of aluminium of 0.047kg placed for sufficient time in a vessel containing boiling water , so that the sphere is at 100C .it is then immediately transferred to 0.14kg copper calorimeter containing 0.25kg of water at 20C .The temperature of water rises and attains a steady state at 23C .Calculate the specific heat capacity of aluminium.

Answer 1

Solution :-

Mass of the aluminium sphere (m1) =0.047kg

Initial temperature of the aluminium sphere = 100oC

Final temperature = 23oC

Change in temperature (DT1)

=100oC-23oC =77oC

Let, specific heat capacity of aluminium = c1.

Amount of heat lost by aluminium

= m1 × c1 × DT1

=0.047kg × c1 × 77oC

Mass of water (m2)

= 0.25kg

Mass of calorimeter (m3)
= 0.14kg

Initial temperature of water and calorimeter

= 20oC

Final temperature

= 23oC

Change in temperature (DT2)

=23oC-20oC = 3oC

Specific heat capacity of water (c2)

= 4.18 × 103J kg-1 oC-1

Specific heat capacity of copper calorimeter (c3)

= 0.386 × 103J kg-1 oC-1

Total amount of heat gained by calorimeter and water

= m2 × c2 × DT2 + m3 × c3 × DT2

  = 0.25kg×4.18 × 103J kg-1 oC-1×3oC+ 0.14kg ×0.386 × 103J kg-1 oC-1×3oC

   = 3297.12J

At steady state, heat lost by aluminium

= heat gained by water+ heat gained by calorimeter.

Therefore,

0.047kg × c1 × 77oC =3297.12J

Therefore, c1

= 911.058 J kg-1 oC-1

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@GauravSaxena01

Answer 2

Given parameters:

Final temperature of the system (after the heated Al is transferred to H2O on Cu calorimeter) = 23 C

m of Al = 0.047 kg

c of Al  = to be calculated

Initial temperature of Al = 100 C

m of H2O = 0.25 kg

m of Cu calorimeter = 0.14 kg

Initial temperature of the system (of Cu calorimeter and H2O before the transfer of the heated Al) = 20C

Standard parameters:

c of H2O = 1.00 kcal/kg C

c of Cu = 0.093 kcal/kg C

Now,

[ Heat lost by Al ] = [Heat gained by H2O] + [Heat gained by Cu calorimeter]

[ (m*c*ΔT) of Al ] = [(m*c*ΔT) of H2O] + [(m*c*ΔT) of Cu calorimeter]

[(0.047kg)*c* (100 C – 23 C) = [(0.25kg)*(1.00 kcal/kg C)*(23 C – 20 C)] + [(0.14kg)*(0.093 kcal/kg C)*(23 C-20 C)]

On solving, we get

c of Al = 0.22 kcal/kg C

∴ Specific Heat Capacity of Al = 0.22 kcal/kg C