Question

A copper ball and an aluminum ball of mass 150 g each are heated to 100°C and then cooled to a temperature of 20°C. The heat lost by the copper ball is 4.6 kJ. The heat lost by the aluminum ball is 10.8 kJ. What is the specific heat of copper, if the specific heat of aluminum is 0.90 J/g°C?

Answer 1

The specific heat of aluminum is actually simply a diversion. Because we can directly compute for the specific heat of copper using the formula:

ΔH = m C ΔT

where ΔH is change in enthalpy or heat lost, m is mass, C is specific heat and ΔT is change in temp

 

4,600 J = 150 g * C * (100 °C - 20°C)

C = 0.38 J/g°C

Answer 2

Answer:

0.383 J/g°C

Explanation:

The heat emitted by the copper ball while cooling down is given by:

$Q=m C_s \Delta T$

where:

m = 150 g is the mass of the ball

Cs = ? is the specific heat of copper

$\Delta T=100^{\circ}C-20^{\circ}C=80^{\circ}C$ is the change in temperature

We know that the heat lost by the copper ball is

$Q=4.6 kJ=4600 J$

So we can re-arrange the formula to find the specific heat capacity:

$C_s = \frac{Q}{m \Delta T}=\frac{4600 J}{(150 g)(80^{\circ} C)}=0.383 J/g^{\circ}C$