# What calorimetry approach is this?

Question:

Find the final tempertature of the system when $5gm$ of steam at $100^\circ C$ is mized with $31.5gm$ ice at $-20^\circ C$.

Constants:
Specific heat of ice and specific heat of steam is $0.5cal/gm/^\circ C$.
Specific heat of water is $1cal/gm/^\circ C$.
Latent heat of fusion is $80cal/gm$
Latent heat of vaporization is $540cal/gm$

My attempt:

I usually approach these problems by setting the final temperature as $T$. Then, I equate heat lost by steam to heat gained by ice. I have been taught this method and I am adept at it.

My problem: The solution given in my workbook is:

Heat released when $1g$ of steam at $100^\circ C$ is cooled to $-20^\circ C$ is $730cal$. Now, this heat is given to be given to $1+6.3g=7.3g$ of ice. This raises the temperature by $10^\circ C$

This is a new method for me, and I am unable to understand it. I wish to know:

How does this method work? In what calorimetry situations can this method be applied?

NOTE: The final answer is $10^\circ C$

• It doesn’t look right to me either. What do you calculate for the final temperature? – Chet Miller Dec 27 '17 at 4:12
• The masses of steam and ice used in the solution seem unrelated to the masses used in the question... – DJohnM Dec 27 '17 at 6:57
• @DJohnM Sorry, I got a typo. Updated it, please have a look now. – Gaurang Tandon Dec 27 '17 at 10:22
• @ChesterMiller It is 10deg celsius – Gaurang Tandon Dec 27 '17 at 10:24
• When you edited it, did you change the mass of ice from 21.5 g to 31.5 g? – Chet Miller Dec 27 '17 at 12:09

The method used in the solution is to:

1. Bring all the constituents (in this case boiling hot steam and sub-zero ice) to a preselected and arbitrary point and calculate the heat flows (in or out) needed to achieve this for each component. In this case, you could pick all liquid water at $0^{\circ}$C.
2. Add up all the individual heat flows to find the net heat flow needed to reach this assumed condition.
3. Finally, realize that there should be zero heat flow. So apply the necessary correction to the single combined mass of water, to find its final temperature and state.

The problem with the method you prefer is that you might need to try many different final states to find the correct answer.

For a very small amount of steam, you might assume that the ice warms up slightly to $T$, while condensing the steam, cooling it from $100^{\circ}$ to $0^{\circ}$, freezing all of it, and cooling it from $0^{\circ}$ to $T$ below zero.

For a larger amount of steam, you might assume that the ice all warms up to $0^{\circ}$ while condensing all the steam, cooling it to $0^{\circ}$, and then freezing $x$ grams of the condensed steam.

For a still larger amount of steam, you might assume that the ice all warms up to $0^{\circ}$ and $x$ grams of it melts, while condensing all the steam and cooling it to $0^{\circ}$.

For a still larger amount of steam you might assume that all the ice warms up to freezing, melts, and warms to $T$, while all the steam condenses and cools to $T$.

And for a really large amount of steam, you might assume that all the ice warms to freezing, melts, warms up to $100^{\circ}$, while $x$ grams of the steam condenses.

Each of these is possible, depending on the amount of ateam, and you have no way of knowing which one to try just by looking at the numbers. You need to try each one. Each produces a different linear equation in one unknown, so each will have a single mathematical solution. But only one of the solutions will be consistent with the assumptions that went into finding it.

You might find values of $x$ that are negative, or too big for the amount of ice or steam actually present. You could assume all liquid water at $T$, and find that $T$ is $125^{\circ}$. You will need to keep trying scenario after scenario, to find the one that makes sense.

• Hi, thanks for your wonderful answer! Could you please add a bit more about how they've manipulated the masses of the substances in their solutions? (it was a typo, i've corrected it now). Notice that the given solution first calculates heat changes caused by 1gram of steam only. – Gaurang Tandon Dec 27 '17 at 10:27
• Hello, your answer is great! But, can you please add more about why they specifically took 1gram of steam only in their solution. Thank you! – Gaurang Tandon Jan 4 '18 at 2:18
• They were confused... – DJohnM Jan 4 '18 at 3:10

There is a difference between the masses in the question and those in the given answer. I assume the masses in the question are correct and those in the given answer are wrong.

In this "method" you imagine that the $5g$ of steam is converted to ice at $-20^{\circ}$C, so that it is the same as the $31.5g$ of ice which is already at $-20^{\circ}$C. The thermal energy extracted is then used to heat $31.5+5=36.5g$ of ice.

This method works because thermodynamic energy changes are conservative like changes in gravitational potential energy. The difference in energy depends only on the initial and final states and not on the path taken. In your method you have two initial states and you assume a temperature for the final state. In this method you convert everything to the same initial state (because it is the same substance) then use the energy extracted to heat it up to the final state.

If you are not happy with this method stick with your usual method, which will give you the same answer.

• Where did the 8 grams of ice come from? – DJohnM Dec 27 '17 at 7:52
• That's a nice way to look at the situation! (I've updated the question with the correct masses, please have a look.) – Gaurang Tandon Dec 27 '17 at 10:28
• Fantastic answer ! first time I understood the calorimeter idea – Buraian Oct 4 '20 at 10:19