Please check my work (ice melting in cup) thanks

[Dars]

Homework Statement

You have 0.9 L of water in an insulated (thermally isolated) cup at 20 ºC. You add 360 g of ice at −20 ºC in the cup to cool the water. Describe the final state of your system in terms of (i) number of moles of ice (ii) number of moles of water and (iii) the final temperature.

Homework Equations

specific heat of liquid water = 75.3JK^-1mol^-1
enthalpy of fusion, 0degC = 6.007kJmol^-1

The Attempt at a Solution

0.9L liquid water = .9kg water: 900g (1mol/18.01g) = 50mol liquid water
360g ice ...........= 20mol ice

The way i am looking at problem is to compare the energy required to melt 20mol ice to the energy released by the water in going from 20degC to 0degC. Is this best way to approach problems such as these?

20mol_ice * 6.007kJmol^-1 = 120.14kJ required to melt 20 moles ice
50mol_liquid water * 75.3JK^-1mol^-1 *20 = 75.3kJ given off by water.

so, using the energy ratio of 75.3:120.14 , the answers are...


(i) 20mol - (75.3kJ/120.14*20mol) = 7.45mol ice

(ii) (20mol - 7.45)_ice - 50mol_water = 62.55mol water

(iii) but i am stuck as to how to find the final temp. I know there must be an equation i am missing that probably makes the fist 2 answers a lot easier also. Please help.. thanks

 

[nate808]

Based on the information given, the final state of the system can be described as follows:

(i) The number of moles of ice will decrease from 20 mol to 7.45 mol, as some of the ice will melt and become liquid water.

(ii) The number of moles of water will increase from 50 mol to 62.55 mol, as some of the ice will melt and become liquid water.

(iii) To find the final temperature, we can use the equation Q = mCΔT, where Q is the heat transferred, m is the mass of the substance, C is the specific heat, and ΔT is the change in temperature.

In this case, the heat released by the water (Q1) is equal to the heat absorbed by the ice (Q2), so we can set them equal to each other:

Q1 = Q2
m1C1ΔT1 = m2C2ΔT2

Substituting in the values we know, we get:

900g * 75.3JK^-1mol^-1 * (Tf - 20ºC) = 360g * 6.007kJmol^-1 + 7.45mol * 75.3JK^-1mol^-1 * (Tf - 0ºC)

Solving for Tf, we get a final temperature of -2.4ºC.

So, the final state of the system will be 7.45 mol of ice, 62.55 mol of liquid water, and a final temperature of -2.4ºC.