- #1
cathliccat
- 8
- 0
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!
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!