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Kaleb
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[SOLVED] Thermo Energy. Iron + Ice
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)
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
All my attempts have met with failure:
1) .11cal/g*C * 50g(x-80C)=.5*y(x-0C) (i don't know the mass of ice so I am 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!
Homework Statement
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)
Homework 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
The Attempt at a Solution
All my attempts have met with failure:
1) .11cal/g*C * 50g(x-80C)=.5*y(x-0C) (i don't know the mass of ice so I am 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!
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