How many g of ice melt before thermal equilibrium is attained

[cathliccat]

The question is "8 grams of water at 100 degrees C are poured into a cavity in a very large block of ice initially at 0 degrees C. How many g of ice melt before thermal equilibrium is attained round off to the nearest whole number?"

As I understnd it I need to:
(heat to change ice to water)+(heat to raise water from 0degrees C to T)=(heat lost by 8g of water cooling from 100degrees C to T)

My book shows the formula for this is:
(mass of the ice*Lf) + (mass of ice*c)=(mass of water*c)(100degrees-T)

So:
m(333kJ/kg)+m(4186J/kg*degC)*(T)=.008kg(4186J/kg*degC)*(100degC-T)

The m is what I'm solving for, what do I do about the T? I know its in equilibrium, but I don't know what it is. I can't solve for both.

Thanks in advance!

 

[HallsofIvy]

"8 grams of water at 100 degrees C are poured into a cavity in a very large block of ice initially at 0 degrees C."

A "very large block of ice". That means that there is sufficient ice that some will still be frozen when equilibrium is reached. The equilibrium temperature is still 0 degrees C. Take T= 0.

 

[M98Ranger]

To solve for the T in this equation, we can use the fact that at thermal equilibrium, the temperature of the ice and water will be the same. This means that T in both the left and right sides of the equation will be equal. So we can substitute T with whatever value we want, as long as it is the same on both sides. Let's choose T = 0 degrees C since that is the initial temperature of the ice.

Plugging in T = 0 degrees C, we get:

m(333 kJ/kg) + m(4186 J/kg*degC)*(0 degC) = .008 kg(4186 J/kg*degC)*(100 degC - 0 degC)

Simplifying this equation, we get:

m(333 kJ/kg) = .008 kg(4186 J/kg*degC)*(100 degC)

Now we can solve for m by dividing both sides by 333 kJ/kg:

m = (.008 kg(4186 J/kg*degC)*(100 degC))/(333 kJ/kg)

m = 0.025 kg = 25 g

Therefore, 25 grams of ice will melt before thermal equilibrium is attained.