Question

How much ice at a temperature of -14.8 ∘C∘C must be dropped into the water so that the final temperature of the system will be 27.0 ∘C∘C ?

Answer 1

Answer:

See explanation below

Explanation:

You are missing some important data, in this case, the mass and temperature of the water. In order to solve for this and explain to you how to do it, I'm going to assume some values for the mass and temperature for water, and then, you only need to replace these values with the values you have to get an accurate answer.

Let's assume the mass of water we have is 0.250 kg of water at 80 °C. Now, 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, let's 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 = 4190 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:

Qw = 0.250 * 4190 * (27 - 80)

Qw = -55,517.5 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 + 14.8)] + (mi * 3.34x10⁵) + [mi * 4190 * (27 - 0)]

Qi = 31,080mi + 3.34x10⁵mi + 113,130mi

Qi = 478,210mi (b)

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

0 = 478,210mi - 55,517.5

55,517.5 = 478,210mi

mi = 55,517.5 / 478,210

mi = 0.1161 kg or 116.1 g of ice is required to drop the temperature of the system to 27.0 °C. Now, remember to replace the value of mass of water and temperature in this procedure to get the real and accuraten answer.