You heat a 5.2 gram lead ball (specific heat capacity = 0.128 joules/g-deg) to 183.°c. you then drop the ball into 34.5 milliliters of water (density 1g/ml; specific heat capacity = 4.184 j/g-deg) at 22.4°c. what is the final temperature of the water when the lead and water reach thermal equilibrium?

  Let T be the final temperature (in °C) to be found:

(0.128 J/g-deg) x (5.2 g) x (183 - T) °C = (121.805 - 0.6656 T) J lost by the Pb
(4.184 J/g-deg) x (34.5 g) x (T - 22.4) °C = (144.348 T - 3233.4) J gained by the H2O
Set the two expressions for the heat lost/gained equal to each other:
121.805 - 0.6656 T = 144.348 T - 3233.4
Solve for T algebraically:
T = 23.1°C