If 85.0 grams of iron (cFe=0.46 J/g°C) at 101.1°C is placed in Coffee Cup Calorimeter,

Question

containing 250.0 grams of water at 25.0°C, and both are allowed to come to thermal equilibrium, what will the final temperature be?

Answer 1

Heat lost by iron must equal heat gained by water

(m_Fe)(C_Fe)(∆T_Fe) = (m_w)(C_w)(∆T_w)

(85.0 g)(0.46 J/gº)(101.1 - Tf) = (250 g)(4.184 J/gº)(Tf - 25)

3953 - 39.1Tf = 1046Tf - 26,150

1085.1Tf = 30103

Tf = 27.7 degrees