Answer to Question #157434 in Physics for Abi

Question

A sample of liquid gold is cooled and allowed to solidify. Calculate the amount of energy in cooling a 0.5kg sample of this gold from 1100 degrees Celsius to 25 degrees Celsius. The melting point of gold is 1063 degrees Celsius, the latent heat of fusion for gold is 65kJ/kg and the specific heat capacity of gold is 130 J/kg ºC. Does this process absorb or release energy? Make a sketch of a temperature-time graph for the reaction.

Answer 1

The specific heat capacity of liquid gold is around 150 kg/(J°C). Consider the processes involved:

1) Liquid cools from 1100°C to 1063°C: $Q_1=c_lm\Delta t_1$

2) Liquid solidifies at 1063°C: $Q_2=m\lambda$

3) Solid gold cools from 1063°C to 25°C: $Q_3=c_sm\Delta t_3$

Since the gold cools down, it gives energy to the environment (the heat is released):

$$Q=Q_1+Q_2+Q_3,\\ Q=c_lm\Delta t_1+m\lambda+c_sm\Delta t_3,\\ Q=0.5[150(1100-1063)+65000+\\[phone])]=102745\text{ J}.$$

Temperature-time graph for the reaction: