Question

help plz ,..,......,....,.........

Answer 1

***heya friend***

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

Cheers!!