What is the specific heat of the unknown sample?

Question

An irregular lump of an unknown metal has a measured density of 3.26 g/mL. The metal is heated to a temperature of 189 °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 34.1 mL, and the temperature is recorded as 40.8 °C. What is the specific heat of the unknown metal sample? Assume no heat is lost to the surroundings.

C=_____

Answer 1

m = Vρ (mass = volume times density)

The volume of that the metal takes up is 34.1 mL - 25.0 mL = 9.1 mL

mass of the metal then is (9.1 mL)(3.26 g/mL) = 29.666 grams

We can also calculate the mass of the water: m = Vρ but you must look up the density of water at 25 °C. I did so and it clocks in at 0.997 g/mL. So, the mass of the water is (25 mL)(0.997 g/mL) = 24.925 grams

Q = mC_pΔT

For the water, we can calculate how much heat was added to the water by the metal as

Q_(water) = m_(water)C_p(water)(ΔT_(water))

We must look up the specific heat (C_p) of water. I did so and got 4.184 J/g˚C

So Q_(water) = (24.925 g)(4.184 J/g˚C)(40.8 - 25.0)°C = 1647.72 joules

The heat lost by the metal is the same bur negative (-1647.73 joules).

For the metal, we can calculate C_p as:

Q_(metal) = m_(metal)C_p(metal)(ΔT_(metal))

C_p = Q_(metal)/(m_(metal)•ΔT_(metal))

C_p = -1647.73/(29.666•(40.8 - 189))

C_p = 0.375 j/(g°C)