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

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Homework Help Overview

The problem involves a calorimetry experiment where a block of ice is added to a calorimeter containing water. The goal is to determine the original mass of the ice block given that only a portion of it melts. The subject area pertains to thermodynamics and phase changes, specifically involving heat transfer and latent heat.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of the heat transfer equation, questioning the use of specific heat versus latent heat for phase changes. There is confusion regarding the temperature change of the ice and the implications for the calculations. Some participants suggest using the latent heat of fusion instead of specific heat for the ice.

Discussion Status

The discussion is active, with participants providing guidance on the correct approach to the problem. There is acknowledgment of errors in unit consistency and the need to clarify the setup of the equations. Multiple interpretations of the problem are being explored, particularly regarding the mass of ice that melts and the corresponding calculations.

Contextual Notes

Participants note the importance of consistent units in calculations and the specific values for specific heats and latent heat. There is an ongoing examination of the assumptions made about the phase change and the initial conditions of the system.

aal0315
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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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