Calculating Final Temperature of Ice-Water Mixture: Grade 10 Physics Question

[smook]

I'm somewhat stumped on this question from my son who's in Grade 10 Physics.

Question:

You put 100g of ice into a glass of 2400g of water at 10C.
What is the final temperature of the water?

Hmmm. Sounds simple,

BUT,

I know,
Q= m(delta)H(fus)

where (delta)H(fus)= 333J/g for water

and

Q=mc(delta)T

where m=mass(g),
c=specific heat of water = 4.18J/gC

BUT we need to find the (delta)T from Tintial to Tfinal.

Where do I start? Or am I missing something here?

 

[Doc Al]

What is the initial temperature of the ice? Let's assume 0°C.

Call the final equilibrium temperature T. The ice will melt then heat up (as water) and the water will cool until the final temperature of everything is T.

The heat needed to raise the temperature of the ice (m) to T is:
Q_ice = mH + mcΔT = mH + mc(T-0)

The heat given off by the water (M) in cooling to T is:
Q_water = McΔT = Mc(10-T)

These heats must be equal. Set up the equation and solve for T.

 

[smook]

Is this not easier...

from Q=mc(delta)T

Tf=Ti-Q/mc

so...

10C-33330Jg/250gx4.18/JC
=6.81C

?

 

[Doc Al]

Is this not easier...

from Q=mc(delta)T

Tf=Ti-Q/mc

so...

10C-33330Jg/250gx4.18/JC
=6.81C

?

It may be easier, but is it correct? I don't really understand what you're doing. What is Q? What is m? Where did you get these values?