What Was the Original Mass of the Ice Block in the Calorimetry Experiment?

AI Thread Summary
The discussion focuses on calculating the original mass of a block of ice in a calorimetry experiment involving an aluminum calorimeter and water. The heat gained by the ice must equal the heat lost by the water and calorimeter, leading to the equation that incorporates the latent heat of fusion for melting ice. Confusion arises regarding the specific heat values and the proper use of units, with a clarification that the specific heats should be consistent in kJ and kg. The final calculations reveal an error due to incorrect unit conversions, which affects the mass result. The participants emphasize the importance of accurate unit handling and the correct interpretation of phase changes in thermal calculations.
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Homework Statement


A block of ice at 0C is added to a 150g aluminum calorimeter cup which holds 210g of water at 12C. If all but 2.0 g of ice melt, what was the original mass of the block of ice


Homework Equations



heat gained by ice = heat lost by water + heat lost by calorimeter

The Attempt at a Solution


so i figure that its:
(m(i)+2.0x10^-3kg)*c(i)*deltaT(i) = (m(w)*c(w)*deltaT(w) +(m(a)*c(a)*deltaT(a), where m is mass, c is the specific heat and T is temperature
but what i don't get is isn't the deltaT(i) going to be zero since the ice starts off zero and the water/ice mixture temperature zero? so then this equation doesn't work because solving for m(i), you can't divide a number by zero?
 
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You need to use the latent heat of fusion to find out how much heat it takes to melt ice.
 
so instead of m(i)*c(i)*deltaT(i) in the above equation, I use m(i)*L, where L is 333 kJ/kg?
 
aal0315 said:
so instead of m(i)*c(i)*deltaT(i) in the above equation, I use m(i)*L, where L is 333 kJ/kg?

Yeah, it's a phase change.
 
so i did what you said and this is what i got
(m(i)+2.0x10^-3kg)L = (m(w)*c(w)*deltaT(w) +(m(a)*c(a)*deltaT(a)
(m(i)+2.0x10^-3)(333) = (210x10^-3*4186*(12-0)) + (150x10^-3*900*(12-0))
m(i) = 36.54kg
this doesn't make sense does it??
 
Wait, shouldn't your specific heats stay at the same numbers?

4.186 J/goC = 4.186 kJ/kgoC
 
what do you mean? little confused ... you said to use the m(i)*L for the left side of the equation and then for the right side, don't i use the specific heats for water and aluminum?
 
aal0315 said:
what do you mean? little confused ... you said to use the m(i)*L for the left side of the equation and then for the right side, don't i use the specific heats for water and aluminum?

You do, but you messed up the numbers. You used 4186 J/kgoC instead of 4.186 kJ/kgoC. So your answer comes out off by a factor of 1000.

Make sure you check that your units are consistent before you answer the question (you used 333 kJ/kg for latent heat so every value should involve kJ and/or kg).

Also, this term is incorrect. (m(i)+2.0x10^-3)(333) All the initial mass BUT 2 g melts so you should subtract.
 
alright .. i get it now .. thanks for the help!
 
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