Kaleb
Feb19-08, 05:48 PM
1. The problem statement, all variables and given/known data
A 50-g chunk of 80 degree C iron is dropped into a cavity in a very large block of ice at 0 degrees C. How many grams of ice will melt? (The specific heat capacity of iron is .11 cal/g*C)
2. Relevant equations
Q = cmdeltaT
Quantity of heat = heat capacity * mass * change in temp
also i know that it takes 80 calories to go just from ice to water
3. The attempt at a solution
All my attempts have met with failure:
1) .11cal/g*C * 50g(x-80C)=.5*y(x-0C) (i dont know the mass of ice so im clueless on how to solve this)
2) .11cal/g*c *50g(0-80C)=.5*y(0-0C) (assuming that the ice will equalize the final temp but it leaves the right equation = 0 which makes it pointless.
There is a re-occurring number that I keep having and its 440 cal/g. Any and all help is appreciated.
I figured it out, my second equation was correct. Thanks though!
A 50-g chunk of 80 degree C iron is dropped into a cavity in a very large block of ice at 0 degrees C. How many grams of ice will melt? (The specific heat capacity of iron is .11 cal/g*C)
2. Relevant equations
Q = cmdeltaT
Quantity of heat = heat capacity * mass * change in temp
also i know that it takes 80 calories to go just from ice to water
3. The attempt at a solution
All my attempts have met with failure:
1) .11cal/g*C * 50g(x-80C)=.5*y(x-0C) (i dont know the mass of ice so im clueless on how to solve this)
2) .11cal/g*c *50g(0-80C)=.5*y(0-0C) (assuming that the ice will equalize the final temp but it leaves the right equation = 0 which makes it pointless.
There is a re-occurring number that I keep having and its 440 cal/g. Any and all help is appreciated.
I figured it out, my second equation was correct. Thanks though!