The Amazing Water Diet - ChemPRIME

The Amazing Water Diet

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back to Heat Capacities

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.

The recommended daily allowance of water is six glasses per day of cold water. This is about 48 fluid ounces, or 1.4 L, at approximately 40oF (~4oC). Our body must use some energy to warm the water to body temperature (37o C).

How much of our daily caloric intake goes into heating the water we drink? We'll calculate that in Example 1, but first we'll need to understand what's meant by the heat capacity of water.

Heat Capacity

When our body supplies heat energy to the water, a rise in temperature occurs (no complicating chemical changes or phase changes take place). The rise in temperature is proportional to the quantity of heat energy supplied. If q is the quantity of heat supplied and the temperature rises from T1 to T2 then

q = C × (T2T1)      (1)


q = C × (ΔT)      (1b)

where the constant of proportionality C is called the heat capacity of the sample. The sign of q in this case is + because the sample has absorbed heat (the change was endothermic), and (ΔT) is defined in the conventional way.

Since the mass of water we drink is variable, it is convenient to note that the quantity of heat needed to raise its temperature is proportional to the mass as well as to the rise in temperature. That is,

q = C × m × (T2T1)      (2)


q = C × m × (Δ T)      (2b)

The new proportionality constant C is the heat capacity per unit mass. It is called the specific heat capacity (or sometimes the specific heat), where the word specific means “per unit mass.”

Specific heat capacities provide a convenient way of determining the heat added to, or removed from, material by measuring its mass and temperature change. As mentioned [|previously], James Joule established the connection between heat energy and the intensive property temperature, by measuring the temperature change in water caused by the energy released by a falling mass. In an ideal experiment, a 1.00 kg mass falling 10.0 m would release 98.0 J of energy. If the mass drove a propeller immersed in 0.100 liter (100 g) of water in an insulated container, its temperature would rise by 0.234oC. This allows us to calculate the specific heat capacity of water:

98 J = C × 100 g × 0.234 oC
C = 4.184 J/goC

At 15°C, the precise value for the specific heat of water is 4.184 J K–1 g–1, and at other temperatures it varies from 4.178 to 4.218 J K–1 g–1. Note that the specific heat has units of g (not the base unit kg), and that since the Centigrade and kelvin scales have identical graduations, either oC or K may be used.

Joule's experiments establish the connection between kinetic and potential energy and heat energy (measured in calories), which is the basis for understanding our metabolic needs.

Example 1: How much food energy is required to raise the temperature of 1,400 mL of water (D = 1.0) from 4.0 oC to 37.0 oC, given that the specific heat capacity of water is 4.184 J K–1 g–1?


q = 4.18 J/goC × 1,400 g × (37.0 - 4.0)
q = 193 000 J or 193 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 193,000 J x (1 calorie / 4.18 J) = 46,200 cal

At first this makes no sense. It appears that 46,200 calories of energy is needed just to heat up the 6 glass of water we drink, but our daily food intake is only 2500 Calories.

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

So 193,000 J x (1 calorie / 4.18 J) x (1 Calorie / 1000 calories) = 46 Cal.

So it does require 46/2500 x 100% or 1.8% of our daily food energy just to heat up the 6 glasses of water to body temperature! That is enough energy to walk nearly 2 miles!

Other foods, and even the air we breath, require different abounts of heat to change their temperature by the same amount. The specific heats of several substances are given below:

Specific heat capacities (25 °C unless otherwise noted)
Substance phase Cp(see below)
air, (Sea level, dry, 0 °C) gas 1.0035
argon gas 0.5203
carbon dioxide gas 0.839
helium gas 5.19
hydrogen gas 14.30
methane gas 2.191
neon gas 1.0301
oxygen gas 0.918
water at 100 °C (steam) gas 2.080
water at 100 °C liquid 4.184
ethanol liquid 2.44
water at -10 °C (ice)) solid 2.05
copper solid 0.385
gold solid 0.129
iron solid 0.450
lead solid 0.127

Example 2: We might breathe around 2 L of cold air per minute at -20 oC on a winter's day, and heat it in our lungs to near 37 oC before exhaling it. How much energy is required for warming inhaled cold air or 3 hours?

Solution: 3 hours x 60 min/hr x 2 L/min = 360 L of air The density of air is about 1.3 g/L at -20 oC m = DV = 1.3 g/L x 360 L = 468 g

q = 1.0035 J/goC × 468 g × (37.0 - (-20.0))
q = 26,800 J or ~27 kJ.

This is only 6-7 dietary Calories.

Electrical Energy Conversion

The most convenient way to supply a known quantity of heat energy to a sample is to use an electrical coil. The heat supplied is the product of the applied potential V, the current I flowing through the coil, and the time t during which the current flows:

q = V × I × t      (2)

If the SI units volt for applied potential, ampere for current, and second time are used, the energy is obtained in joules. This is because the volt is defined as one joule per ampere per second:

1 volt × 1 ampere × 1 second = 1\begin{matrix}\frac{\text{J}}{\text{A s}}\end{matrix} × 1 A × 1 s = 1 J

EXAMPLE 3: An electrical heating coil, 230 cm3 of water, and a thermometer are all placed in a polystyrene coffee cup. A potential difference of 6.23 V is applied to the coil, producing a current of 0.482 A which is allowed to pass for 483 s. If the temperature rises by 1.53 K, find the heat capacity of the contents of the coffee cup. Assume that the polystyrene cup is such a good insulator that no heat energy is lost from it.

Solution The heat energy supplied by the heating coil is given by

q = V × I × t = 6.23 V × 0.482 A × 483 s = 1450 V A s = 1450 J


q = C × (T2T1)

Since the temperatue rises, T2 > T1 and the temperature change ΔT is positive:

1450 J = C × 1.53 K

so that

\begin{matrix}C=\frac{\text{1450 J}}{\text{1}\text{.53 K}}=\text{948 J K}^{-\text{1}}\end{matrix}

Note: The heat capacity found applies to the complete contents of the cup-water, coil, and thermometer taken together, not just the water.

As discussed in other sections, an older, non-SI energy unit, the calorie, was defined as the heat energy required to raise the temperature of 1 g H2O from 14.5 to 15.5°C. Thus at 15°C the specific heat capacity of water is 1.00 cal K–1 g–1. This value is accurate to three significant figures between about 4 and 90°C.

If the sample of matter we are heating is a pure substance, then the quantity of heat needed to raise its temperature is proportional to the amount of substance. The heat capacity per unit amount of substance is called the molar heat capacity, symbol Cm. Thus the quantity of heat needed to raise the temperature of an amount of substance n from T1 to T2 is given by

q = C × n × (T2T1)      (4)

The molar heat capacity is usually given a subscript to indicate whether the substance has been heated at constant pressure (Cp)or in a closed container at constant volume (CV).

EXAMPLE 4: A sample of neon gas (0.854 mol) is heated in a closed container by means of an electrical heating coil. A potential of 5.26 V was applied to the coil causing a current of 0.336 A to pass for 30.0 s. The temperature of the gas was found to rise by 4.98 K. Find the molar heat capacity of the neon gas, assuming no heat losses.

Solution The heat supplied by the heating coil is given by

   q = V × I × t
= 5.26 V × 0.336 A × 30.0 s
= 53.0 V A s
= 53.0 J

Rearranging Eq. (4), we then have

\begin{matrix}C_{m}=\frac{q}{n\text{(T}_{\text{2}}-\text{T}_{\text{1}}\text{)}}=\frac{\text{53}\text{.0 J}}{\text{0}\text{.854 mol }\times \text{ 4}\text{.98 K}}=\text{12}\text{.47 J K}^{-\text{1}}\text{ mol}^{-\text{1}}\end{matrix}

However, since the process occurs at constant volume, we should write/>

CV = 12.47 J K–1 mol–1


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