Expert: Josh - 7/12/2007

Question
Which would require more heat, melting 500 g of 0 C ice or turning 500 g of 100 C water into steam?

Hints:

The formulas used to calculate the amount of heat absorbed during melting or vaporization are:

amount of heat absorbed in fusion = (mass of the material) x (latent heat of fusion)
Q = mLf

and

amount of heat absorbed in vaporization = (mass of the material) x (latent heat of vaporization)
Q = mLv

where Q is the amount of heat in calories;
          m is the mass of the material in grams;
          Lf and Lv is the latent heat of fusion and the latent heat of vaporization measured in calories/gram.

Note:
The latent heat of fusion of water: Lf = 79.8 cal/gram
The latent heat of vaporization of water: Lv = 539.6 cal/gram


Answer
Hi Chris,

We need to compare the amount of heat required to melt 0.5 kg of ice at zero degree Celsius, to the amount of heat required to evaporate 0.5 kg of water into steam.

The former is described by:
 H2O(s) + Q_fusion -> H2O(aq)
The latter is described by:
 H2O(l) + Q_evaporation -> H2O(g)

where Q_fusion = m*Lf, Q_evaporation = m*Lv.

Since the mass is the same (not talking about volume here, clearly water has a different density to ice), we can ignore the mass altogether.

Instead of asking whether Q_fusion > Q_evaporation,
i.e., m*Lf > m*Lv, this is the same as checking whether Lf > Lv.

The latent heat of fusion, and vaporization are both measured in cal/gram. This provides a straight-forward comparison.

In fact, Lf=79.8 << Lv=539.6. It requires much more energy to evaporate water at 100 degree, than melting ice at 0 degree.

Ans: Turning 500 g of 100 C water into steam requires more heat.