Some people believe that if by placing a cold spoon in a cup of hot coffee, it will cool it enough to drink it comfortably. Let’s test this. If you have a silver spoon that has been chilled at T= 10.0 °C (let’s say mass = 100.0 g, assume 100% Ag, specific heat =0.235 J/g °C), and you place it in a 240. mL cup of coffee that is at T= 90.0 °C (a typical temperature at a McDonald’s restaurant).
What will the final temperature of the coffee? (Food for thought, if you spill this coffee, will it be hot enough to give you 3rd-degree burns?)

Explanation:

There are two heat transfers involved: the heat gained by the spoon and the heat lost by the coffee.

According to the Law of Conservation of Energy, energy can neither be destroyed nor created, so the sum of these terms must be zero.

Let the spoon be Component 1 and the coffee be Component 2.

Data:  

For the spoon:

For the coffee:

Calculations

1. The relative temperature changes

2. Final temperature of coffee

14 days is the half-life of phosphorus-32

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3