﻿﻿ The Amazing Ice Diet - ChemPRIME

# The Amazing Ice Diet

We typically require 2500 Calories per day in food to supply our energy needs. The Calorie is a unit of energy that is produced in our bodies through oxidation of food by oxygen in the air we breath. The energy is equal to that obtained by burning the food in oxygen (which is how the caloric value is obtained). If we don't use the food in our diet to produce energy, it gets stored (mostly as fat). If we include ice in our diet as suggested by some Extreme Diets, could the fat be avoided by using food energy to melt the ice?

Ever since the revolutionary work of Joseph Black (1728-1799), we know that a huge amount of energy is absorbed when ice melts, even though the temperature remains at 0 oC throughout the process.

If this is true, could we just eat ice cubes to "burn Calories"? Suppose we consume 6 ice cubes, 3 cm on an edge in a day. How much of our daily caloric input is required to melt the ice? We'll be able to answer that question (in Example 1), but first we need to understand the meaning of "heat of fusion".

## Heat of Fusion

When heat energy is supplied to ice at say, -10oC, we find that the temperature climbs steadily until the melting point (0 oC) is reached and the first signs of liquid formation become evident. Thereafter, even though we are still supplying heat energy to the system, the temperature remains constant as long as both liquid and solid are present. Only when the last vestiges of the solid have disappeared does the temperature start to climb again.

This macroscopic behavior demonstrates quite clearly that energy must be supplied to ice (or any solid) in order to melt it. On a microscopic level melting involves separating molecules which attract each other. This requires an increase in the potential energy of the molecules, and the necessary energy is supplied by the heating coil. The kinetic energy of the molecules (rotation, vibration, and limited translation) remains constant during phase changes, because the temperature does not change.

The heat energy which ice absorbs when it melts is called the enthalpy of fusion or heat of fusion and is usually quoted on a molar basis. (The word fusion means the same thing as “melting.”) When 1 mol of ice, for example, is melted, we find from experiment that 6.01 kJ are needed. The molar enthalpy of fusion of ice is thus +6.01 kJ mol–1, and we can write

H2O(s) → H2O(l)      (0°C)      ΔHM = 6.01 kJ mol–1

Selected molar enthalpies of fusion are tabulated below. Solids like ice which have strong intermolecular forces have much higher values than those like CH4 with weak ones.

Example 1: How much food energy (in Calories) is required to melt six ice cubes which are 3 cm on an edge?

Solution V = (3 cm)3 = 27 cm3

The density of ice is about 0.92 g/cm3

m = V x D = 27 cm3 x 0.92 g/cm3 = 220 g

q = n x ΔHM = 220 g (1 mol / 18 g) x 6.01 kJ mol–1= 72.5 kJ

To convert this energy to the US/British unit, we use a conversion that comes from the specific heat of water in those units, 1.0 calorie/g oC:

4.18 J = 1 calorie

So

72 500 J x (1 calorie / 4.18 J) = 17 300 cal

At first this makes no sense. It appears that 17 300 calories of energy is needed just to melt the 6 ice cubes, but our typical daily food intake is only about 2500 Calories.

The confusion lies in the definition of a dietary "Calorie" (with a capital "C"). 1 Calorie = 1000 calories

So

72 500 J x (1 calorie / 4.18 J) x (1 Calorie / 1000 calories) = 17 Cal.

So it does require 17/2500 x 100% or 0.7% of our daily food energy just to heat up the 6 ice cubes water to body temperature! That is enough energy to walk a good part of a mile!

## Enthalpy of Vaporization

Energy is also consumed by evaporation of perspiration, and evaporation and exhalation of water in each breath. Evaporation of perspiration removes heat during exercise to cool our bodies, but we may not recognize the amount of water we exhale. If we breath about 10,000 L of air per day, and the air is about 4% water vapor, we exhale about 400 L of water vapor or roughly 16 mol of water per day. How much of our daily caloric intake is reqired just to evaporate that water (ignoring the energy removed by evaporating perspiration)?

To answer that question, we need to look into the "heat of vaporization":

When a liquid is boiled, the variation of temperature with the heat energy supplied is similar to that found for melting. When heat is supplied at a steady rate to a liquid at atmospheric pressure, the temperature rises until the boiling point is attained. After this the temperature remains constant until the enthalpy of vaporization has been supplied. Once all the liquid has been converted to vapor, the temperature again rises. In the case of water the molar enthalpy of vaporization is 40.67 kJ mol–1. In other words

H2O(l) → H2O(g)      (100°C)      ΔHM = 40.67 kJ mol–1

Molar Enthalpies of Fusion and Vaporization of Selected Substances.

 Substance Formula ΔH(fusion) / kJ mol1 Melting Point / K ΔH(vaporization) / kJ mol-1 Boiling Point / K (ΔHv/Tb) / JK-1 mol-1 Neon Ne 0.33 24 1.80 27 67 Oxygen O2 0.44 54 6.82 90.2 76 Methane CH4 0.94 90.7 8.18 112 73 Ethane C2H6 2.85 90.0 14.72 184 80 Chlorine Cl2 6.40 172.2 20.41 239 85 Carbon tetrachloride CCl4 2.67 250.0 30.00 350 86 Water* H2O 6.00678 at 0°C, 101kPa6.354 at 81.6 °C, 2.50 MPa 273.1 40.657 at 100 °C, 45.051 at 0 °C, 46.567 at -33 °C 373.1 109 n-Nonane C9H20 19.3 353 40.5 491 82 Mercury Hg 2.30 234 58.6 630 91 Sodium Na 2.60 371 98 1158 85 Aluminum Al 10.9 933 284 2600 109 Lead Pb 4.77 601 178 2022 88

Example 2 How many Calories are required to evaporate 16 mol of water per day (roughly the amount we exhale)?

Solution

q = n x ΔHM = 16 mol x 40.67 kJ mol–1 = 651 kJ

651 kJ x (1 Cal / 4.18 kJ) = 156 Cal

It appears that a substantial amount of energy goes into evaporating water!

Heat energy is absorbed when a liquid boils because molecules which are held together by mutual attraction in the liquid are jostled free of each other as the gas is formed. Such a separation requires energy. In general the energy needed differs from one liquid to another depending on the magnitude of the intermolecular forces. We can thus expect liquids with strong intermolecular forces to have larger enthalpies of vaporization. The list of enthalpies of vaporization given in the table bears this out.

Two other features of the table deserve mention. One is the fact that the enthalpy of vaporization of a substance is always higher than its enthalpy of fusion. When a solid melts, the molecules are not separated from each other to nearly the same extent as when a liquid boils. Second, there is a close correlation between the enthalpy of vaporization and the boiling point measured on the thermodynamic scale of temperature. Periodic trends in boiling point closely follow periodic trends in heat of vaporiation. If we divide the one by the other, we find that the result is often in the range of 75 to 90 J K–1 mol–1. To a first approximation therefore the enthalpy of vaporization of a liquid is proportional to the thermodynamic temperature at which the liquid boils. This interesting result is called Trouton’s rule. An equivalent rule does not hold for fusion. The energy required to melt a solid and the temperature at which this occurs depend on the structure of the crystal as well as on the magnitude of the intermolecular forces.