How much ice (in grams) would have to melt to lower the temperature of 353 mLmL of water from 26 ∘C∘C to 6 ∘C∘C? (Assume the density of water is 1.0 g/mLg/mL.)

mi = 0.097 kg or 97 g.

Explanation:

In order to do this, let's remember that the water and ice are exerting heat and the sum of these heats must be zero always:

Q = Qw + Qi = 0 (1)

In the case of the heat exerted by the ice, remember that the ice is passing through several stages:

First, it's temperature changes but not it's phase, it remains solid.

Second, it's when the ice begins to melt and ends when it's totally melted.

And third, the change of temperature of water when the ice melted. So for these three stages, the heat of ice can be calculated with the following expression:

Qi = (mi * Ci * ΔTi) + (mi * Lf) + (mi * Cw * ΔTm) (2)

The values of Ci, Lf and Cw are tabulated and reported data, and these are the following:

Ci: Specific heat of ice = 2100 J/kg °C

Cw: Specific heat of water = 4186 J/kg °C

Lf: heat of fusion of water = 3.34x10⁵ J/kg

For the case of heat of water it's just the specific heat of water, it's mass and the difference of temperature and the expression is:

Qw = mw * Cw * ΔT (3)

Now, let's calculate firt the heat of water. As the density of water is 1 g/mL, the volume and it's mass are the same, so we have 353 g of water. Using this value, and the expression above, let's calculate the heat for water:

Qw = (353/1000) * 4186 * (6 - 26)

Qw = - 29,553.16 J (a)

We have the heat of water, now, let's calculate heat of ice in function of the mass of ice:

Qi = [mi * 2100 * (0 - 26) ] + (mi * 3.34x10⁵) + [mi * 4190 * (6 - 0) ]

Qi = - 54,600mi + 3.34x10⁵mi + 25,140mi

Qi = 304,540mi (b)

Finally, replace (a) and (b) in equation (1) to solve for mass of ice:

0 = 304,540mi - 29,553.16

29,553.16 = 304,540mi

mi = 29,553.16/304,540

mi = 0.097 kg of ice is required to lower the temperature to 6 °C.