Question

A 70.0g piece of metal at 80.0c is placed in 100g of water at 22.0c contained in a calorimeter like that shown in figure. The metal and water come to the same temperature at 24.6c. How much heat did the metal give up to the water? What is the specific heat of the metal?

Answer 1

Heat lost by the metal = Heat gain by the water

or, - Q_m = Q_w

Let us say that the specific heat of the metal = Z

Specific heat of water = 4.186 J/g ^oC

Heat lost by the metal, Q_m = mass of the metal x specific heat of the metal x (final temperature - initial temperature)

                                           = 70.0 x Z x (24.6 - 80)

                                   = - 3878Z J

Heat gained by water, Q_w = mass of the water x specific heat of water x (final temperature - initial temperature)

                                          = 100 x 4.186 x (24.6 - 22.0)

                                          = 1088.36 J

                                  

As,

- Q_m = Q_w

Therefore,

- (- 3878Z) = 1088.36

or, 3878Z = 1088.36

or, Z = 1088.36/3878

        = 0.281 J/g ^oC

Therefore, the specific heat of the metal = 0.281 J/g ^oC

And, the amount heat given up by the metal = the amount of heat heat gained by the water = 1088.36 J