In: Chemistry
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.
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