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If I have cup of water at room temperature (say, $25^\circ$C). What would be the resultant temperature if I pour another cup of same amount of water at $100^\circ$C to it? Is it simply $\frac{25+100}{2}$?

If yes, what if I pour a cup of n times amount of water at $100^\circ$C to it? Is it simply the sum of all temperature (25+100) divided by the total amount of water $(n+1)$ i.e. $\frac{25+100n}{n+1}$?

I am no expert in Physics. I am just thinking of a way to get a cup of water at roughly, say, $40^\circ$C without using any thermometer. I know I can obtain water at room temp and $100^\circ$C easily. But dont know how to get my target temp by mixing water at these 2 different temp.

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Heat transfers from hotter water to cooler water until temperatures equalize.

If mass and temperature of the hotter water are $m_H$ and $T_H$, mass and temperature of the cooler water are $m_C$ and $T_C$, specific heat capacity of water is $c$, and equilibrium temperature is $T$, then heat released by hotter water $Q_1$ is

$$Q_1 = c\ m_H (T_H-T)$$

while the heat absorbed by cooler water $Q_2$ is

$$Q_2 = c\ m_C (T-T_C)$$

Since $Q_1 = Q_2$ you can easily obtain equilibrium temperature as

$$T = \frac{m_H T_H + m_C T_C}{m_C+m_H}$$

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    $\begingroup$ As he is talking about cups, it should be in terms of volume and not mass, right? So I think the answer should include the thermal expansion coefficient somewhere. $\endgroup$
    – Bernhard
    Commented Apr 26, 2012 at 10:43
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    $\begingroup$ In principle you are right. However the thermal expansion coefficient is not temperature independent, and even if it would be temperature independent you end up with exponential change of volume. I think it is much better idea to use data for water's density and recalculate volumes into masses... You think that I should include that in the text? $\endgroup$
    – Pygmalion
    Commented Apr 26, 2012 at 11:13
  • $\begingroup$ For water at $100^\circ C$, do we have to take into account latent heat? $\endgroup$
    – leongz
    Commented Apr 26, 2012 at 11:46
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    $\begingroup$ @leongz Only if you use vapour instead of liquid water. $\endgroup$
    – Pygmalion
    Commented Apr 26, 2012 at 12:11
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    $\begingroup$ @Pygmalion: This only works due to the constant specific heat of water, which is a reasonable approximation. $\endgroup$
    – Ron Maimon
    Commented Apr 28, 2012 at 5:44

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