Calorimeter question for chemistry

Question

A hot lump of 30.5 g of aluminum at an initial temperature of 69.2 degrees C is placed in a 50.0 mL H_2O initially at 25.0 degrees C and allowed to reach thermal equilibrium. What is the final temperature of the aluminum and water, given the specific heat of aluminum is 0.903 J/(g• degrees C)? Assume no heat is lost to surroundings

Answer 1

Heat lost by hot Al MUST equal heat gained by cool water.

q = mC∆T

q = heat

m = mass

C = specific heat

∆T = change in temperature (Tf = final temp)

Heat lost by hot Al = q = (30.5 g)(0.903 J/gº)(69.2º - Tf)

Heat gained by H2O = q = (50.0 g)(4.184 J/gº)(Tf - 25.0º)

Setting these equal to each other and solving for Tf, we have...

(30.5 g)(0.903 J/gº)(69.2º - Tf) = (50.0 g)(4.184 J/gº)(Tf - 25.0º)

1906 - 27.5Tf = 209.2Tf - 5230

236.7Tf = 7136

Tf = 30.1ºC

(be sure to check all of the calculations)