Cooling Water with Ice#

TL;DR#

One ice cube cools two ounces of water to just above freezing temperature.

To cool 12 oz of water from 20°C to just above freezing, approximately 88.9 grams of ice is needed. A standard ice cube (approximately 16 grams) can be used as an approximation, requiring about six ice cubes.

Background#

I have this lovely home water carbonation machine. It dispenses water in 6 oz increments, and 12 oz is my sweet spot. The effectiveness of water absorption of carbon dioxide depends on water temperature. The colder the better. The machine doesn’t cool the water itself and has little insulation in the water tank (which I think is a design flaw). My fridge produces ice cubes, though. Following is a school way of figuring out how much ice I must put into a tank for a glass of bubbly water.

Interactive form#

Amount of water: 12 oz

Room temperature: 72 °F

Ice cube weight: 16 g

Number of ice cubes needed: 6

Time needed: 5 minutes

Physics#

To calculate the amount of ice needed to cool 12 oz (355 mL) of water from 20°C to 0°C, we’ll use the principle of energy conservation. The heat lost by the water must equal the heat gained by the ice.

Assumptions#

Ice melting point is 0°C.
The specific heat capacity of water is 4.18 J/g°C.
The latent heat of fusion for ice (the energy required to melt ice at 0°C) is 334 J/g.
The density of water is 1 g/mL, so 355 mL of water is 355 grams.

Step 1: Calculate the heat energy lost by the water#

\[Q_{\text{water}} = m_{\text{water}} \times c_{\text{water}} \times \Delta T_{\text{water}}\]

Where:

\(Q_{\text{water}}\) is the heat lost by the water,
\(m_{\text{water}} = 355 \, \text{g}\),
\(c_{\text{water}} = 4.18 \, \text{J/g°C}\),
\(\Delta T_{\text{water}} = 20 \, \text{°C} - 0 \, \text{°C} = 20 \, \text{°C}\).

\[Q_{\text{water}} = 355 \, \text{g} \times 4.18 \, \text{J/g°C} \times 20 \, \text{°C} = 29,678 \, \text{J}\]

Step 2: Calculate the amount of ice needed#

Since the ice will absorb heat and melt, the heat required to melt \(m_{\text{ice}}\) grams of ice is given by:

\[Q_{\text{ice}} = m_{\text{ice}} \times L_{\text{fusion}}\]

Where:

\(Q_{\text{ice}}\) is the heat absorbed by the ice,
\(L_{\text{fusion}} = 334 \, \text{J/g}\) (latent heat of fusion).

Since the heat lost by the water equals the heat gained by the ice:

\[Q_{\text{water}} = Q_{\text{ice}}\]

\[29,678 \, \text{J} = m_{\text{ice}} \times 334 \, \text{J/g}\]

\[m_{\text{ice}} = \frac{29,678 \, \text{J}}{334 \, \text{J/g}} \approx 89 \, \text{g}\]

My ice cubes weigh 16 grams on average, so I need about six ice cubes.

Step 3: Calculate the amount of time needed#

For a rough estimate, assuming the heat transfer rate \(R\) is about 100 W (100 J/s, a typical estimate for moderate convective conditions):

\[t = \frac{Q_{\text{ice}}}{R}\]

\[t = \frac{29,678 \, \text{J}}{100 \, \text{J/g}} \approx 297 \text{seconds} \approx 5 \text{minutes}\]

Calculation flaws#

The formula above doesn’t take into consideration:

Room air warming up water as the ice melts.
The volume of water increases from the melted ice. With 89 grams of molten ice, I need only 9 ounces of water to get 12 ounces of ice-cold water.
The convection varies significantly as the water temperature changes. 100 J/s might be true for room temperature, but it evidently takes more than 10 minutes for all ice cubes to melt completely. But this is fine, water is cold enough for carbonation purposes.