What ice pack formula can hold the largest energy difference?

Question

I recently bought some Freez Pak brand ice packs. The label advertises: "Colder Than Ice". I realize now that it's just a stupid marketing claim, but at first, it got me wondering if they had some special formula in them.

Some possibilities of different formulations I thought of:

1. They put denatured alcohol inside so it could get cold without freezing, but the paks do indeed freeze solid.

2. They put salt and water inside to lower the freezing point. (?! It was just a thought!)

3. It's just regular water inside because the label is simply just marketing.

Number 3 is the most likely scenario because simple water is the cheapest.

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Creating My Own Ice Packs

The principle here is a substance that can hold the biggest difference in energy. Temperature is irrelevant because given 12 hours, anything in my freezer will be at the same -15$^\circ$C.

Let's assume when the ice pack warms up to 5$^\circ$C it is no longer useful for keeping food fresh. I will be filling 1 liter bottles.

Thermal Energy Formula:

$E$ = (Specific Heat kJ/kg $^\circ$C) $\cdot$ ($\Delta$T $^\circ$C) $\cdot$ (1 liter) $\cdot$ (Density kg/L)

Water Only:

Water's Specific Heat = 4.2 kJ/kg$^\circ$C

Water's Density = 1 kg/L

• $E_1$ = 5$^\circ$C water to just-before-freezing energy
• $E_2$ = Energy opposite of melting ice
• $E_3$ = Just-before-freezing to -15$^\circ$C ice energy
• $E_{Total}$ = $E_1$ + $E_2$ + $E_3$

$E_1$ = 4.2 kJ/kg$^\circ$C $\cdot$ ( 5$^\circ$C - 0$^\circ$C ) $\cdot$ 1 kg = 21 kJ

$E_2$ = 334 kJ/kg $\cdot$ 1 kg = 334 kJ

$E_3$ = 4.2 kJ/kg$^\circ$C $\cdot$ [ 0$^\circ$C - (-15$^\circ$C) ] $\cdot$ 1 kg = 63 kJ

$E_{Total}$ = 21 kJ + 334 kJ + 63 kJ = 418 kJ

That phase change gave me a ton of energy!

Alcohol Only:

Unfortunately, alcohol won't pass its freezing point of -114$^\circ$C in my kitchen fridge. :/

Alcohol's (ethyl's) Specific Heat = 2.4 kJ/kg$^\circ$C

Alcohol's Density = .8 kg/L

$E_{Total}$ = 2.4 kJ/kg$^\circ$C $\cdot$ [ 5$^\circ$C - (-15$^\circ$C) ] $\cdot$ .8 kg = 38.4 kJ

Wow! That would be a worthless ice pack!

Other Potential Formulas:

1. Salt Water; Salt has a low specific heat by itself, but I don't know what happens when mixed with water.
2. Sugar Water; Same as salt.
3. Ammonia has a higher specific heat, but won't go through that wonderful phase change.
4. Glycol!

Glycol:

Glycol has a freezing point at -13$^\circ$C! My freezer can do that.

Glycol's Specific Heat = ~.5 (? I couldn't find this one, but estimated based off graphs)

Glycol's Density = 1.1 kg/L

• $E_1$ = 5$^\circ$C glycol to just-before-freezing energy
• $E_2$ = Energy opposite of melting glycol
• $E_3$ = Just-before-freezing to -15$^\circ$C glycol ice energy
• $E_{Total}$ = $E_1$ + $E_2$ + $E_3$

$E_1$ = .5 kJ/kg$^\circ$C $\cdot$ ( 5$^\circ$C - 13$^\circ$C ) $\cdot$ 1.1 kg = 9.9 kJ

$E_2$ = 181 kJ/kg $\cdot$ 1.1 kg = 199 kJ

$E_3$ = .5 kJ/kg$^\circ$C $\cdot$ [ -13$^\circ$C - (-15$^\circ$C) ] $\cdot$ 1.1 kg = 1.1 kJ

$E_{Total}$ = 9.9 kJ + 199 kJ + 1.1 kJ = 210 kJ

Nope! That's still half the capacity of water.

Questions:

What have I done wrong in my math?

What materials and/or properties have I overlooked?

Answer 1

I will not answer your question directly; only give you some tools that should help you answer the question (in practice) yourself.

To focus the attention, find below a typical heating/cooling diagram for a frozen pure substance.

The vertical axis marked $T$ represents temperature (in degrees Celsius).

Three significant temperatures are indicated on the $T$ axis:

$T_c$: this is the temperature of the cold pack while in your freezer.

$T_f$: this is the melting (fusion) point of the material in the pack.

$T_H$: this is the temperature after the pack has been carrying out its function for some time. Often it will be room temperature.

The horizontal axis marked $H$ represents Enthalpy. Enthalpy is what is known in physics as Heat Energy. $H_1$, $H_2$ etc are various heat energy contents of the pack as it slowly heats up from $T_c$ to $T_H$.

You’re quite correct that the best pack (there are other, practical criteria, of course) would be the one that on heating would extract the highest amount of Enthalpy from its surroundings (typically a cooled picnic box or such like) to heat up from $T_c$ to $T_H$. So let’s analyse this.

The overall Enthalpy the pack has to absorb is simply $\Delta H = H_4 - H_1$. We can calculate this as follows.

The heating diagram is divided in three areas marked $I$, $II$ and $III$.

Area $I$: the frozen pack is heated from $T_c$ to $T_f$. The Enthalpy change for area $I$ is given by:

$\Delta H_I=mC_s(T_f – T_c)$ where $m$ is the mass (in $kg$) of the pack and $C_s$ is the specific heat capacity of the solid pack material (in $J kg^{-1} K^{-1}$).

Area $II$: in this area the pack is melting (during melting $T$ stays constant at $T_f$). The Enthalpy change for area $II$ is given by:

$\Delta H_{II}=mL_f$, with $L_f$ the latent heat of fusion (in $J kg^{-1}$). For melting $L_f$ is a positive value.

Area $III$: the molten pack is heated from $T_f$ to $T_H$. The Enthalpy change for area $III$ is given by:

$\Delta H_{III}= mC_l(T_H – T_f)$ where $C_l$ is the specific heat capacity of the liquid pack material (in $J kg^{-1} K^{-1}$).

We can now add it all up: $\Delta H =\Delta H_I + \Delta H_{II} + \Delta H_{III}$.

Or:

$\Delta H = m[C_s(T_f – T_c) + L_f + C_l(T_H – T_f)]$

We can now conclude a few things, in order to maximise $\Delta H$:

1. High mass $m$ is desirable.
2. Large difference $T_H-T_c$ is desirable.
3. Both $C_s$ and $C_l$ should be as large as possible.
4. Large value of $L_f$ also increases $\Delta H$.

This general treatment was for a pack filled with a single pure substance but can as an approximation also be used for mixtures of substances, by adjusting the masses, latent heats and specific heat capacities.