Question

An irregular lump of an unknown metal has a measured density of 5.71 g/mL. The metal is heated to a temperature of 155 °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 32.9 mL, and the temperature is recorded as 47.4 °C. What is the specific heat of the unknown metal sample? Assume no heat is lost to the surroundings.

c=

Jg⋅°C

Answer 1

volume of unknown = 32.9 mL - 25.0 mL

= 7.9 mL

Since density of unknown = 5.71 g/mL,

mass = density * volume

= 5.71 g/mL * 7.9 mL

= 45.1 g

Since density of water is 1 g/mL,

m(water) = 25.0 g

T(water) = 25.0 oC

C(water) = 4.184 J/goC

m(unknown) = 45.1 g

T(unknown) = 155.0 oC

C(unknown) = to be calculated

We will be using heat conservation equation

use:

heat lost by unknown = heat gained by water

m(unknown)*C(unknown)*(T(unknown)-T) = m(water)*C(water)*(T-T(water))

45.1*C(unknown)*(155.0-47.4) = 25.0*4.184*(47.4-25.0)

4852.76*C(unknown) = 2343.04

C(unknown)= 0.4828 J/goC

Answer: 0.483 J/goC