Question

An irregular lump of an unknown metal has a measured density of 5.63 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 31.9 mL, and the temperature is recorded as 39.5 °C. What is the specific heat of the unknown metal sample? Assume no heat is lost to the surroundings.

Answer 1

Calculate mass of unknown metal.

Volume of water before addition of unknown metal = 25.0 ml

Volume of water after addition of unknown metal = 31.9 ml

volume of unknown metal = Volume of water after addition of unknown metal - Volume of water before addition of unknown metal

volume of unknown metal = 31.9 ml - 25.0 ml = 6.9 ml

Density of unknown metal = 5.63 g/ ml

We have relation, density = Mass / volume

Mass of unknown metal = density volume

Mass of unknown metal = 5.63 g/ ml 6.9 ml = 38.8 g

Mass of water = 0.99707 g /ml 25.0 ml = 24.93 g ( density of water at 25 ⁰ C = 0.99707 g/ml)

Assume no heat is lost to surrounding. Therefore, Heat lost by metal is absorbed by water.Hence, we can write

q metal + q water = 0

Heat absorbed or emitted by any substance is given as q = m x C x (T final - T initial )

Where q is a heat absorbed or emitted, m is a mass of a body, C is a specific heat capacity of a body, T is a temperature of a body.

We have , final temperature of both water & metal = 39.5 ⁰ C , Initial temperature of water = 25.0 ⁰ C , Initial temperature of metal = 167  ⁰ C & specific heat capacity of water = 4.184 J / g ⁰ C

[ 38.8 g C ( 39.5 ⁰ C - 167 ⁰ C)] _metal + [ 24.93 g 4.184 J / g ⁰ C ( 39.5 ⁰ C - 25.0 ⁰ C)] _water = 0

[ 38.8 g C (- 127.5  ⁰ C )] _metal + 1512.4 J = 0

[ 38.8 g C (- 127.5  ⁰ C )] _metal = - 1512.4 J

C = - 1512.4 J / ( 38.8 g ( -127.5  ⁰ C )

C = 0.3057 J / g ⁰ C

C = 0.306 J / g ⁰ C

ANSWER : Specific heat capacity of unknown metal = 0.306 J / g ⁰ C