Water Ball

What will be the final, equilibrium temperature of the ball and the water?

A ball of aluminum [c = 0.897 J/g°C] has a mass of 100 grams and is initially at a temperature of 150°C. This ball is quickly inserted into an insulated cup containing 100 ml of water at a temperature of 15.0°C.

Heat lost by Aluminium = Heat gained by water.
‘T’ = final temp.
Al: 100g x 0.897J/g/°C x (150 – T) ΔT.
= 13,455 – 89.7T
H2O: 100g x 4.18J/g/°C x (T – 15) ΔT.
= 418T – 6,270.

13,455 – 89.7T = 418T – 6,270.
13,455 + 6,270 = 418T + 89.7T.
19,725 = 507.7T
T = 19,725 / 507.7 = 38.85°C final temp. (39°C)

Water Balls