When a small ice cube at 10°C is placed into a cup of water at room temperature, both the warming of the ice and the melting of ice into water will occur. However, the impact of each process on the overall cooling of the liquid water can differ significantly.
To start, let’s consider the specifics: the specific heat capacity of ice is 2.09 J/g°C. This means that it takes 2.09 joules of energy to raise the temperature of one gram of ice by 1°C. When the ice warms up from 10°C to 0°C, it absorbs energy from the surrounding water. If we assume the ice cube has a mass of, say, 10 grams, it will absorb:
Q_ice warming = mass x specific heat capacity x temperature change
Q_ice warming = 10 g x 2.09 J/g°C x (0°C – 10°C) = -209 J (energy absorbed by ice)
Next, we consider the heat of fusion of ice, which is given as 6.02 kJ/mol. In order to melt the ice completely, it will require additional energy. Here we have to convert this heat of fusion into joules:
Heat of fusion = 6.02 kJ/mol x 1000 J/kJ = 6020 J/mol
Given the molar mass of water is approximately 18 g/mol, the melting of 10 g of ice would involve:
Q_melting = (10 g / 18 g/mol) x 6020 J/mol ≈ 3344 J
Now, the total energy absorbed by the ice consists of both the energy needed to warm the ice and the energy required to melt it:
Total energy absorbed by ice = Q_ice warming + Q_melting = -209 J + 3344 J ≈ 3135 J
Comparatively, this total is significantly larger than the initial energy that the water could lose due to its own cooling. Therefore, in this scenario, the melting of ice plays a much larger role in cooling the liquid water compared to the warming of the ice itself. The substantial energy absorbed during the phase change from solid to liquid (melting) effectively cools the water more than any initial temperature change of the ice.