Specific Heat of Unknown Metal Sample

Question

An irregular lump of an unknown metal has a measured density of 5.25 g/mL. The metal is heated to a temperature of 167 °C and placed in a graduated cylinder filled with 25.0 mL of water at 25.0 °C. After the system has reached thermal equilibrium, the volume in the cylinder is read at 31.0 mL, and the temperature is recorded as 46.3 °C. What is the specific heat of the unknown metal sample? Assume no heat is lost to the surroundings.

Answer 1

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

First, we will want to find the mass of the metal. This can be done by knowing the density (5.25 g/ml) and the volume of the irregular shaped lump. The volume will be equal to the volume of water that it displaces, in this case, that will be 31.0 ml - 25.0 ml = 6.0 mls.

Mass of metal = 5.25 g/ml x 6.0 mls = 31.5 g

Now we can return to our heat equation to find the heat lost by the metal and the heat gained by the water.

For the metal:

q = mC∆T = (31.5 g)(C)(167º - 46.3º) = (31.5 g)(C)(120.7º) = heat lost by metal

For the water (assuming a density of 1 g/ml and a specific heat of 4.184 J/g/deg):

q = mC∆T = (25.0 g)(4.184 J/g/deg)(46.3 - 25.0 deg) = (25.0 g)(4.184 J/g/deg)(25.8 deg) = 2699 J = heat gained by water

Heat lost by the metal must equal the heat gained by the water, so we can write the following:

(31.5 g)(C)(120.7 deg) = 2699 J

C = 2699 J/(31.5 g)(120.7 deg)

C = 0.710 J/g/deg = specific heat of the metal