If 15g of a metal at 150C with a heat capacity of 0.81 J g/C is added to 100.g of water at 25C at 4.1J/gC, what temperature will the combined system arrive at?

Answer 1

Heat lost by the metal = heat gained by the water

heat lost by the metal = mC∆T = (15 g)(0.81 J/g/deg)(150 - Tf) where Tf = final temperature

heat gained by the water = mC∆T = (100 g)(4.184 J/g/deg)(Tf - 25)

(15 g)(0.81 J/g/deg)(150 - Tf) = (100 g)(4.184 J/g/deg)(Tf - 25)

1823 - 12.15 Tf = 418.4Tf - 10460

430.6Tf = 12283

Tf = 28.5ºC