Question

An aluminium sphere of mass 0.047 kg is heated
100°C. It is dropped in a copper calorimeter of
0.14 kg, containing 0.25 kg of water at 20°C. The
temperature of water rises to a steady state at 23°C.
Calculate the specific heat of aluminium. Specific heat
of water = 4.18 x 10 J kg-1 °C-1, specific heat of
copper = 0.386 x 103 J kg-1•c-1.​

Answer 1

Answer:

911.058

Explanation:

mass of Al (m1) = 0.047kg

change in temperature (CT1) = 100 - 23 = 77

mass of water (m2) = 0.25kg

mass of calorimeter (m3) = 0.14kg

change in temp. (CT2) = 23 - 20 = 3

Q = heat

c = specific heat

c of aluminum (c1) = ?

c of water (c2) = 4.18 x 10

c of copper (c3) = 0.386 x 103

formula  Q=mcCT

              Q = [ (m2*c2*CT2) + (m3*c3*CT2)]

              Q = 3297.12 J

solve for the specific heat of aluminum.

Q = m1*c1*CT1

3297.12 = 0.047*c1*77

c1 = 911.058