Given,
final temp is 33°C
for metal ball,
mass m = 200 gm = 0.2 kg
difference in temp = 50 °C
specific heat capacity s = ?
heat lostr by metal ball = Q1 = msdt
= 0.2 * s * 50
= 10 s........ (i)
Again for water,
mass m = 300 gm= 3°C
specific heat capacity s = 4200 J/kg°C
Amount of heat gained by water Q2 = msdt
= 0.3 * 4200 * 3
= 3780 J ..... (ii)
By the principle of calorimetry
heat lost by metal ball = heat gained by water
Q1 = Q2
10 S = 3780
S = 378 J/ Kg°C