# What ice pack formula can hold the largest energy difference?

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.

## 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?

• The post is kind of hard to read because of the lack of mathjax use and all the bold etc. The users here are expert physicists; there's no need to bold important parts of the question. If the question is hard for expert physicists to understand without bold then it needs to be clarified anyway. – DanielSank Aug 28 '15 at 0:32
• @DanielSank I'm just a regular guy and I think this question could appeal to other non-experts. I agree that it could look better, but I don't now how to do that. I'm cool with someone editing it though. Just tell me how. – Zach Mierzejewski Aug 28 '15 at 0:36
• Interesting paper with some other formulations mentioned: pharmoutsourcing.com/Featured-Articles/… – BowlOfRed Aug 28 '15 at 0:37
• @ZachMierzejewski Hi Zach, I edited out the bold, but I left you to practice your Mathjax on your own post :) 5 minutes of Google should sort out any math formatting questions you might have. If your pack is pure water and you bought it with that claim, colder than ice, get your money back:) Best of luck with it anyway. – user81619 Aug 28 '15 at 1:09
• I'm not really expert in such things but for maximum "cold" you want it to freeze solid, or most of it to freeze anyway, as the un-freezing takes heat. A 32 degree ice cube absorbs much more heat than the equivalent amount of 32 degree water. But the alcohol might soften and make it more slushy so it can shape around something while cold (nobody would want to apply a rock hard ice pack to their skin). I suspect it's a trade-off between maintaining some softness/flexibility and cold retention. The goal probably isn't maximum heat absorption. By weight, pure water might be best at that. – userLTK Aug 28 '15 at 1:15

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.

• Thank you for showing me the real names of all the properties and their proper symbols. I didn't know that there was a different $C_s$ and $C_l$ depending on the phase, but that makes sense because those are variable with temperature too. I upvoted, but haven't accepted yet because, as you said, it's not a direct answer. – Zach Mierzejewski Aug 28 '15 at 3:04
• Thanks. Check the heat capacity of ice for example: engineeringtoolbox.com/ice-thermal-properties-d_576.html and compare to water. – Gert Aug 28 '15 at 13:20