Question

A 72.5 gram piece of magnesium is heated to a temperature of 98.90ºC. The metal is then dropped into 40.0 grams of water at a temperature of 18.50ºC inside a perfect calorimeter. Calculate the final temperature of the water in the calorimeter. The specific heat capacity for magnesium is Cp = 1.020 J/g·ºC.

Tfinal = ºC

Answer 1

m(water) = 40.0 g

T(water) = 18.5 oC

C(water) = 4.184 J/goC

m(metal) = 72.5 g

T(metal) = 98.9 oC

C(metal) = 1.02 J/goC

T = 23.9 oC

We will be using heat conservation equation

Let the final temperature be T oC

use:

heat lost by metal = heat gained by water

m(metal)*C(metal)*(T(metal)-T) = m(water)*C(water)*(T-T(water))

72.5*1.02*(98.9-T) = 40.0*4.184*(T-18.5)

73.95*(98.9-T) = 167.36*(T-18.5)

7313.655 - 73.95*T = 167.36*T - 3096.16

T= 43.1388 oC

Answer: 43.1 oC