Question

An irregular lump of an unknown metal has a measured density of 5.07 g/mL. The metal is heated to a temperature of 157 °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 30.5 mL, and the temperature is recorded as 45.5 °C. What is the specific heat of the unknown metal sample? Assume no heat is lost to the surroundings.

Answer 1

Solution : -

Volume of metal = 30.5 ml – 25.0 ml= 5.5 ml

Mass of metal= 5.5 ml *5.07 g per ml =27.885 g

Lets calculate the amount of heat absorbed by water

q= m*c*delta T

= 25 g * 4.184 J per g C * (45.5 C – 25 C)

= 2144.3 J

Same amount of the heat is loss by the metal

So using the value of the heat lets calculate the specific heat of the metal

q=m*c*delta T

c=q/ m*delta T

= 2144.3 J /27.885 g * (157 C – 45.5 C)

= 0.6897 J per g C

So the specific heat of the metal is 0.6897 J/ g C