Predict thermal equilibrium temps

Question

A 14.69-g block of solid titanium at 65.34 °C is immersed in a 23.08-g pool of liquid ethanol with a temperature of 14.71 °C. When thermal equilibrium is reached, what is the temperature of the titanium and ethanol?

Specific heat capacities: titanium = 0.523 J/g °C; ethanol = 2.44 J/g °C

Answer 1

Heat lost by titanium MUST equal heat gained by cold ethanol

heat = mC∆T

heat lost by titanium = (14.69 g)(0.523 J/gº)(65.34 - Tf)

heat gained by ethanol = (23.08 g)(2.44 J/gº)(Tf - 14.71)

Setting these two equal we have..

(14.69 g)(0.523 J/gº)(65.34 - Tf) = (23.08 g)(2.44 J/gº)(Tf - 14.71)

Solve for Tf (final temperature)...

502 - 7.68Tf = 56.3Tf - 828

1330 = 63.98Tf

Tf = 20.8ºC