Question a5378

Mar 19, 2017

You will need a minimum of 119 grams of ice.

Explanation:

There are three events occurring here, and each has its own energy factor:

The ice must warm to 0°C. We will call the energy for this change ${E}_{1}$

The ice must melt at 0°C (without further change in temperature). We will call the energy for this change ${E}_{2}$

The water must cool from 10°C to 0°C. We will call the energy for this change ${E}_{3}$

The first two of these will require an absorption of energy. The third (which is occurring simultaneously with the other two) will supply all this energy.

Conservation of energy requires that ${E}_{3} = {E}_{1} + {E}_{2}$

For the two effects that involve temperature changes, the equation is

$E = m \times c \times \Delta t$

where $m$ is the mass (of ice or water), $c$ is the specific heat of the substance (you look this up) and $\Delta t$ is the temperature change.

For ice c=2.11 J/(g °C)

So ${E}_{1} = m \times 2.11 \times 10 = 21.1 m$

For water c=4.18 J/(g °C)#

This means ${E}_{3} = 500 \times 4.18 \times 20 = 41 800 J$

Finally, to find ${E}_{2}$, we use the heat of fusion of water (or the heat of melting of ice if you prefer to think of it that way). This is 334 L/g.

So ${E}_{2} = m \times \left(334\right)$

Putting it all together, remembering that ${E}_{3} = {E}_{1} + {E}_{2}$ we get

$41 800 = 21.1 m + 334 m = 351.1 m$

resulting in $m = 41 \frac{800}{351.1} = 119 g$