Question 8.41

[Michelle_Tan_1G]

An ice cube is added to liquid water at given quantities and initial temperatures, and we are asked to find the final temperature. How do I factor in the enthalpy required to melt the ice cube?

So far I have:

qsys = qsurr
qsolid = qliquid
H(enthalpy of fusion) + nC∆T = nC∆T

Is this the correct way to structure my answer?

[Rosaline_Chow_2L]

Hi,

So first, I began by calculating the heat required to melt the ice, which is q1 = 50g/(18.02g/mol) * 6.01kJ/mol = 16,700J. Then I calculated the heat required to get the ice, now in liquid state, to get to the final temperature, which is q2 = (50g)(4.184J/g*°C)(Tf-0°C) or simplifying it, 209.2Tf. Adding these two together gives you qsys = 16,700J + (209.2J/°C)Tf.

Afterwards, I calculated qsurr = (400g)(4.184J/g*°C)(Tf-45°C), simplifying it to (1673.6J/°C)Tf - 75,312J.

Now qsys + qsurr = 0, so qsys = -qsurr.
16,700J + (209.2J/°C)Tf = -[(1673.6J/°C)Tf - 75,312J]
(1882.8J/°C)Tf = 58,612J
Tf =31°C

This is just the way I do it, but I hope it helps!

[Michelle_Tan_1G]

That makes sense, thank you so much!