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

Click For Summary
SUMMARY

The original mass of the ice block in the calorimetry experiment was calculated to be 36.54 kg, but this result was incorrect due to unit inconsistencies and misunderstanding of the phase change. The correct approach involves using the latent heat of fusion (L = 333 kJ/kg) for the ice and ensuring that specific heats are consistently applied in kJ and kg. The heat gained by the ice must equal the heat lost by the water and the calorimeter, taking into account that only 2.0 g of ice remains unmelted.

PREREQUISITES
  • Understanding of calorimetry principles
  • Knowledge of latent heat of fusion
  • Familiarity with specific heat capacities of water and aluminum
  • Ability to perform unit conversions in thermodynamic equations
NEXT STEPS
  • Review the concept of latent heat and its application in phase changes
  • Learn about specific heat capacities of various materials
  • Practice solving calorimetry problems involving phase changes
  • Explore unit conversion techniques in thermodynamic calculations
USEFUL FOR

Students studying thermodynamics, educators teaching calorimetry concepts, and anyone interested in understanding heat transfer in phase changes.

aal0315
Messages
41
Reaction score
0

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?
 
Physics news on Phys.org
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!
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
668
Replies
12
Views
1K
  • · Replies 17 ·
Replies
17
Views
6K
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
7K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 4 ·
Replies
4
Views
5K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
2
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K