Calorimetry with ice and water

In summary, when adding a block of ice to a calorimeter cup of water at 12C, the heat gained is equal to the heat lost by the calorimeter, water, and ice.
  • #1
aal0315
41
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?
 
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  • #2
You need to use the latent heat of fusion to find out how much heat it takes to melt ice.
 
  • #3
so instead of m(i)*c(i)*deltaT(i) in the above equation, I use m(i)*L, where L is 333 kJ/kg?
 
  • #4
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.
 
  • #5
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??
 
  • #6
Wait, shouldn't your specific heats stay at the same numbers?

4.186 J/goC = 4.186 kJ/kgoC
 
  • #7
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?
 
  • #8
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.
 
  • #9
alright .. i get it now .. thanks for the help!
 

1. How does calorimetry with ice and water work?

In calorimetry with ice and water, a known amount of ice is added to a known volume of water at a known temperature. The ice will absorb heat from the water, causing it to melt. By measuring the change in temperature of the water, the amount of heat released by the water can be calculated.

2. What is the purpose of using ice in calorimetry?

The ice serves as a way to measure the heat capacity of the water. By measuring the amount of heat required to melt the ice, the heat capacity of the water can be determined. This is an important factor in many thermodynamic calculations.

3. How is the heat capacity of ice and water related?

The heat capacity of ice and water are related because they are both forms of the same substance - H2O. However, the heat capacity of ice is lower than that of water, meaning that it requires less heat to change the temperature of ice compared to water.

4. Can calorimetry with ice and water be used to calculate the energy content of a substance?

Yes, calorimetry with ice and water can be used to calculate the energy content of a substance. By measuring the change in temperature of the water and using the known heat capacity of water, the amount of heat released can be calculated. This value can then be used to determine the energy content of the substance.

5. What are some common applications of calorimetry with ice and water?

Calorimetry with ice and water is commonly used in chemistry and physics experiments to determine the heat capacity of substances, calculate energy content, and study thermodynamic processes. It is also used in industries such as food and beverage production to measure the energy content and quality of products.

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