How Much Ice Is Needed to Cool Water by a Given Temperature?

To find how much ice is needed, compare the heat lost by the warm water with the heat absorbed by the ice. The ice first melts, then the meltwater warms to the mixture’s final temperature. Both energy changes matter.

The formula below assumes the water and ice are mixed in an insulated container, the ice starts at 0 °C, and the final mixture remains above 0 °C. It neglects heat absorbed by the container and heat exchange with the surroundings.

Heat-balance formula

Let $m_w$ be the initial mass of water, $T_i$ its initial temperature, and $\Delta T$ the desired drop. The final temperature is

$$T_f = T_i - \Delta T.$$

If the water cools but does not freeze, it releases

$$Q_{\text{water}} = m_w c_w \Delta T,$$

where $c_w$ is the specific heat capacity of liquid water, approximately $4.18\ \text{kJ/(kg·°C)}$ near ordinary temperatures.

Each kilogram of ice at 0 °C absorbs energy in two stages. It takes about $L_f = 334\ \text{kJ/kg}$ to melt, then the melted water absorbs $c_w T_f$ as it warms from 0 °C to the final temperature. Thus, for an ice mass $m_i$,

$$Q_{\text{ice}} = m_i(L_f + c_w T_f).$$

Equating the heat lost and gained gives

$$\boxed{m_i = \frac{m_w c_w \Delta T}{L_f + c_w(T_i-\Delta T)}}.$$

Use consistent units: for example, masses in kilograms, specific heat in kJ/(kg·°C), and latent heat in kJ/kg. A temperature difference of 1 °C has the same numerical size as a difference of 1 K.

Worked example

Suppose you have 1.00 kg of water at 25 °C and want to cool it to 10 °C. The desired drop is $\Delta T=15\ ^\circ\text{C}$, so the water releases

$$Q_{\text{water}} = (1.00)(4.18)(15) = 62.7\ \text{kJ}.$$

Each kilogram of ice at 0 °C absorbs about 334 kJ while melting. The resulting water then warms from 0 °C to 10 °C, absorbing another

$$(4.18)(10) = 41.8\ \text{kJ/kg}.$$

So each kilogram of ice absorbs about $334+41.8=375.8\ \text{kJ}$. The required ice mass is

$$m_i = \frac{62.7}{375.8} = 0.167\ \text{kg}.$$

You need about 167 g of ice. This calculation includes the meltwater, which becomes part of the final mixture.

If the ice starts below 0 °C

Ice from a freezer may need to warm to 0 °C before it can melt. If its starting temperature is $T_{i,\text{ice}}<0\ ^\circ\text{C}$, include that extra energy. Using $c_{\text{ice}}\approx2.1\ \text{kJ/(kg·°C)}$, the heat absorbed per kilogram of ice is approximately

$$c_{\text{ice}}(0-T_{i,\text{ice}}) + L_f + c_wT_f.$$

Replace the denominator in the boxed formula with this expression. For example, ice starting at −10 °C needs about $2.1\times10=21\ \text{kJ/kg}$ just to warm to its melting point, in addition to the energy for melting and warming the meltwater.

When the simple formula does not apply

The target final temperature must be above 0 °C for the assumption that all the ice melts and the mixture stays liquid to hold. If the calculated target is exactly 0 °C, the ideal heat balance can bring the mixture to that temperature. A target below 0 °C cannot be reached by simply adding ice to liquid water at ordinary pressure: water and ice in equilibrium are at about 0 °C. A different setup, such as adding salt or using a colder environment, changes the problem.

For a real container, some energy may warm the container, and heat may enter from the room. Those effects usually mean you need more ice than the ideal estimate. If they are significant, include them as additional heat terms in the balance.

Common mistakes

• Counting only the energy to melt the ice. Meltwater must also warm to the final temperature, unless the final temperature is 0 °C.
• Warming the meltwater to the initial water temperature. In the ideal final mixture, it warms only to $T_f$.
• Using the initial temperature in the meltwater term. The correct term is $c_wT_f$, because the meltwater starts at 0 °C.
• Mixing units. If you use $c_w$ in kJ/(kg·°C), use $L_f$ in kJ/kg and mass in kilograms.
• Ignoring the container or surroundings. The formula is an ideal insulated-system estimate; real heat exchange can change the amount required.