Lead in Calorimeter Physics/Chemistry Question

A 169.0 g piece of lead is heated to 81.0 degrees C and then dropped into a calorimeter containing 628.0 g of water that initally is at 18.0 degrees C. Neglecting the heat capacity of the container, find the final equilibrium temperature (in degrees C) of the lead and water.

Thanks for any help

Specific Heat capacity of Lead = 130 J/Kg C
Specific Heat capacity of water = 4180 J /Kg C

Let equilibrium temp be θ degrees C
Heat lost by lead = heat gained by water
Heat is calculated from mass x specific heat capacity x temp. change
δ
0.169 x 130 x (81 - θ) = 0.628 x 4180 x (θ - 18)
1779.6 - 21.97 θ = 2625 θ - 47251
49031 = 2647 θ
θ = 18.5 °