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

Can u please help me with this question ?

Hi Nikesh!

Mass of the aluminium sphere (m1) =0.047kg

Initial temperature of the aluminium sphere = 100^oC

Final temperature = 23^oC

Change in temperature (DT1) =100^oC-23^oC =77^oC

Let, specific heat capacity of aluminium = c1.

Amount of heat lost by aluminium = m1 × c1 × DT1 =0.047kg × c1 × 77^oC

Mass of water (m2) = 0.25kg

Mass of calorimeter (m3) = 0.14kg

Initial temperature of water and calorimeter = 20^oC

Final temperature = 23^oC

Change in temperature (DT2) =23^oC-20^oC = 3^oC

Specific heat capacity of water (c2) = 4.18 × 10³J kg^{-1 o}C^-1

Specific heat capacity of copper calorimeter (c3) = 0.386 × 10³J kg^{-1 o}C^-1

Total amount of heat gained by calorimeter and water = m2 × c2 × DT2 + m3 × c3 × DT2

  = 0.25kg×4.18 × 10³J kg^{-1 o}C^-1×3^oC+ 0.14kg ×0.386 × 10³J kg^{-1 o}C^-1×3^oC

   = 3297.12J

At steady state, heat lost by aluminium = heat gained by water+ heat gained by calorimeter.

Therefore,

0.047kg × c1 × 77^oC =3297.12J

Therefore, c1 = 911.058 J kg^{-1 o}C^-1

Cheers!!