Solving a basic calorimetry problem

Question

A 53.7g sample of aluminum, which has a specific heat capacity of 0.897 J* g^-1 * C^-1 , is put into a calorimeter (see sketch at right) that contains 200.0 g of water. The temperature of the water starts off at 16.0 C. When the temperature of the water stops changing it's 19.9 C. The pressure remains constant at 1 atm. Calculate the initial temperature of the aluminum sample. Be sure your answer is rounded to 2 significant digits.

Answer 1

The heat lost by the aluminum sample plus the heat gained by the water equals zero.

Q_Al + Q_w = 0.  Therefore,

Q_Al = - Q_w

Q_Al  = sp. Heat Al x mass Al x (Tf – To_Al)

Q_w = sp. Heat water x mass water x (Tf -To_w)

4.186 J/°C g x 200.00g x (19.9 – 16.0) = 3265 J gained by water

0.897 J/°C g x 53.7g x (19.9 – To_Al) = 959 J – 48.2 xTo_Al J lost by Al.

3265 J = -(959 J + 48.2 To_Al   Solve for To_Al

To_Al =  88 °C

Check Math.