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I have a question in calorimetry from an old competitive exam. The question is:

The temperature of $100$ grams of water is to be raised from $24 ^\circ$C to $90 ^\circ$C by adding steam to it. Find the mass of steam required.

I tried attempting the question by assuming that the added steam would convert back to water and thus lose an amount of heat calculated by the latent heat of vaporization. Additionally, to cool from 100 degrees to 90 degrees, the 'watered steam' will contribute some additional heat. I equated this to the heat gained by the 100 grams of water to reach 90 degrees from 24 degrees. But I am not getting the right answer!

I would appreciate any help on this matter. Thank you :)

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I believe this problem is under-defined. The steam can be any temperature over $100^\circ \:\mathrm{C}$ so ignoring plasma, a single water molecule could do assuming a truly absurd temperature (energy) for it. – Brandon Enright May 21 '13 at 4:41
Is it a reasonable assumption that steam is at $100 ^\circ$C? This is a question from a competitive exam. Perhaps there are some obvious assumptions that I am not aware of. Would you kindly tell me the answer assuming steam is at 100 degree? Thanks – Isomorphism May 21 '13 at 5:53
Do either you or the exam find 12 grams as the answer? – DJohnM May 21 '13 at 6:09
up vote 1 down vote accepted

Your method seems correct. Here are the details:

$Q_{Heat Water}=m\times c\times (T_2-T_1)=100 \times 1 \times 66 = 6600$ calories

From 1 gram of steam, $$Q=L_v+c\times(100-T_2)=540+1 \times 10=550$$ Therefore, grams of steam needed $=\frac{6600}{550}=12$ grams

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Oh! I essentially did the same thing. But I made a calculation mistake. Thank you for your answer :) – Isomorphism May 21 '13 at 6:37

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