Thermodynamics of Mixing Ice and Water in a Glass Container

[haydn]

Homework Statement

M_I kilograms of ice is placed into a M_G kilogram glass container holding M_W kilograms
of water. The water and the glass are initially at 25 degrees C. If the ice came from a freezer at - 15 degrees C, what is the final temperature of the drink assuming that there is not enough ice to freeze the water?
You are given:
L_V for water
L_F for water
Specific heat of water is C_W
Specific heat of ice is C_I
Specific heat of glass is C_G

Homework Equations

Q=mc$$\Delta$$T
Q= + or - mL

The Attempt at a Solution

I know that when the ice is placed in the glass container holding water, all three will transfer energy and reach an equilibrium temperature. I know to set up the equation using:

Q_cold = -Q_hot

With the glass and water being the "hot" and the ice being the "cold." The loss of heat of the hot is equal to the gain in heat of the cold.

I know Q_hot will be equal to mc$$\Delta$$T for the glass and the water.

What I'm not sure about is Q_cold... I know it will increase in temperature so I need to include an mc$$\Delta$$T, but how do I know whether or not it is going to pass 0 degrees C and undergo a phase change to liquid?

Thanks.

 

[davieddy]

The assumption in the question implies that all the ice gets melted,
since the final temperature > 0 degrees

 

[haydn]

Ok, so will the Q_cold expression include mL and mc$$\Delta$$T or just mL? And please explain why...

Thank you.

 

[davieddy]

Heat gained by what starts as ice = heat lost by drink + glass