(a) Convert to SI unit: 1 J/g °F

(b) Why does 1 g of water at 0°C have 336 J more heat energy than 1 g of ice at 0°C?

(a) 1 J/g °F = 1800 J / (kg K)

(b) 336J of energy is used to increase the potential energy due to an increase in intermolecular separation, so the latent heat of ice is 336 J.

Explanation:

(a) ∵ 1 gram = 0.001 kg

$\because 1 °F = \dfrac{5}{9} K$

Now convert: $1~\text{J}/{\text{g °F}} \\[1em] = \dfrac{1\text{J}}{0.001 \text{kg} \times \dfrac{5}{9} \, \text{K}} \\[1em] = \dfrac{1}{0.001 \times \dfrac{5}{9}} ~~\text{J/(kg K)}\\[1em] = \dfrac{9}{0.005} = 1800~ \text{J/(kg K)}$

(b) At 0 °C, ice and water are at the same temperature, but not at the same energy state. To convert 1 g of ice at 0 °C into 1 g of water at 0 °C, it must absorb latent heat.

This latent heat of fusion for water is: L_f = 336 J/g

So, to melt 1 g of ice: Q = mL = 1 g x 336 J/g = 336 J

Thus, 1 g of water at 0 °C has 336 J more energy than ice at the same temperature because that’s the energy absorbed during the phase change.