# Thermodynamics - Latent Heat Problem (finding mass)

• Galois_
In summary, the equation you're trying to use to solve this problem is: mi = [mw(cw)(ti-tf)]/[Lfi + (Tf - Ti)]This equation is incorrect because you can't add the latent heat of ice ('LFi') to a temperature difference (Tf - Ti).

#### Galois_

Homework Statement
A 0.19 kg water is initially at 23C. How much ice (in grams), initially at -20C must be mixed into it, to achieve a final temperature of 3C with all the ice melted. Neglect the heat capacity of the container. Use cw = 1 cal/gK, ci = 0.48 J/gK, and Lfi = 79.6 cal/g.
Relevant Equations
mi = [mw(cw)(ti-tf)]/[Lfi + (Tf - Ti)]
Can someone walk me through this? Our teacher didn't bother to meet us synchronously ever since and I'm kinda stumped on solving this. The equation I used for this problem is:

mi = [mw(cw)(ti-tf)]/[Lfi + (Tf - Ti)]
mi = [190(1)(23-3)]/[79.6+(3+20)] = 37.04g

mi = mass of ice
mw = mass of water
cw = specific heat of water
ci = specific heat of ice
Lfi = latent heat of ice

I'm not sure if that's right or not because I couldn't really understand anything from the module he gave us. Thanks!

You have to decompose the problem into 3 steps (because what happens at 0°C?).

Galois_ said:
Use cw = 1 cal/gK, ci = 0.48 J/gK, and Lfi = 79.6 cal/g.
You are mixing calories and joules as energy units. That's a recipe for confusion and disaster! Express all values using only one energy unit (preferably (joules).

What values will you use?

Galois_ said:
Relevant Equations:: mi = [mw(cw)(ti-tf)]/[Lfi + (Tf - Ti)]
The equation is (very) wrong for a number of reasons.

You can't add the latent heat of ice ('LFi') to a temperature difference (Tf - Ti). It makes no sense because they have different dimensions - in the same way that adding 5grams to 2 seconds makes no sense.

Galois_ said:
Can someone walk me through this?
We can try to guide you, so you do it for yourself. Check the rules: https://www.physicsforums.com/threads/homework-help-guidelines-for-students-and-helpers.686781/

If you were asked to fnd an expression for the amount of energy needed to raise m grams of ice from -20ºC to 3ºC how would you do it? (See @DrClaude's Post #2 hint).

DrClaude said:
You have to decompose the problem into 3 steps (because what happens at 0°C?).
Steve4Physics said:
If you were asked to fnd an expression for the amount of energy needed to raise m grams of ice from -20ºC to 3ºC how would you do it? (See @DrClaude's Post #2 hint).
Sorry, I tried reading more about the topic because the videos and supplementary readings our teacher gave us really didn't help. I couldn't really think anything but this:

1. Ice starts to melt (absorbing heat from the water until melting temp @ 0C)
2. Once that all of the ice has melted, the water from the ice will rise in temperature until...
3. The system reaches equilibrium

I'm still struggling with forming the whole equation for the problem, though. I'm really sorry I don't understand how it works.

Steve4Physics said:
You are mixing calories and joules as energy units. That's a recipe for confusion and disaster! Express all values using only one energy unit (preferably (joules).
I'll be using cal. If I did it correctly, 0.48 J/gK = 0.1147 cal/g, right?
Steve4Physics said:
You can't add the latent heat of ice ('LFi') to a temperature difference (Tf - Ti). It makes no sense because they have different dimensions - in the same way that adding 5grams to 2 seconds makes no sense.
Figured. I just used the equation that was used in another problem and just solved for m. Thanks for pointing that out, though. Will try to do dimensional analysis more frequent in future problems.

Galois_ said:
I'll be using cal. If I did it correctly, 0.48 J/gK = 0.1147 cal/g, right?
No.

The specific heat capacity of ice is about half that of liquid water. So it should be about 0.5 cal/(gK). You need to check your original question/data.

And you mean 'cal/gK' not 'cal/g'.
________________

Initially, think of the ice and the warm water as two separate items.

To get the m grams of frozen water (i.e. the ice) from -23ºC [correction] -20ºC to 3ºC there are three separate stages:
a) Heat the solid ice from -23ºC to 0ºC
b) Melt the ice at 0ºC
c) Warm up the resulting water from 0ºC to 3ºC.

You calculate these three separate energies (giving expressions which include ‘m’). The total energy needed is the sum of these three expressions.

Then ask yourself ‘where has this energy come from?'.

Edit - typo's corrected.

Last edited:
Galois_ said:
1. Ice starts to melt (absorbing heat from the water until melting temp @ 0C)
This is not right. Assuming heat energy is being transferred to the ice, it starts to melt at 0 oC and can stay at 0 oC until all of it is melted.

It takes 334 J of heat energy to melt one gram of ice, a process that occurs at a constant temperature of 0 oC.

Galois_ said:
I just used the equation that was used in another problem
Not a good idea unless the other problem was the same as the one you're trying to solve. In the course you're taking you're being evaluated on your ability to apply the physics to situations, not your ability to copy formulas.

Calorimetry problems are easy if you think in terms of how much heat is needed to do what and where it must come from. In this problem you need to convert ##m## kg of ice to water at 3 °C.
To do that you need
(a) heat ##Q_1## to raise the temperature of the ##m## kg of ice from -20 °C to 0 °C;
(b) heat ##Q_2## to convert ##m## kg of ice at 0 °C to ##m## kg of water at 0 °C;
(c) heat ##Q_3## to raise the temperature of the ##m## kg of the melted ice from 0 °C to 3 °C;

All that heat must come from the initial 0.19 kg of water in the bath which loses heat ##Q_4## while its temperature drops from 23 °C to 3 °C. This is an amount for which you can easily find a numerical value because you have all the numbers. Since no heat is lost, you need to set it equal to the sum of ##Q_1##, ##Q_2## and ##Q_3## because the Joules that are lost by the water bath are gained by the ice. Needless to say, you must first find expressions for these in terms of ##m## and then solve for ##m##.

DrClaude