how to find the specific heat

Question

A 63.4-gram piece of metal is heated to 90.9 °C. It is placed into 50.0 g water with an initial temperature of 20.1 °C. After stirring, the final water temperature is 38.8 °C. What is the specific heat (J/g⋅°C) of the metal? Answer to 3 decimal places with no units.

Answer 1

The heat lost by the hot metal MUST EQUAL the heat gained by the cooler water.

q = mC∆T where q = heat; m = mass; C = specific heat; ∆T = change in temperater

heat lost by hot metal = q = mC∆T = (63.4 g)(C)(90.9 - 38.8) = (63.4)(C)(52.1) = 3303.14 C

heat gained by water = q = mC∆T = (50.0 g)(4.184 J/gº)(38.8 - 20.1) = 209.2 x 18.7 = 3912.04

Setting the two equal to each other and solving for C of the metal, we have...

3303.14 C = 3912.04

C = 1.184 J/gº

Answer = 1.184