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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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1 Answer

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

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

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

while the heat absorbed by cooler water is

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

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

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

$c = 4200$ kJ/kgK is specific heat capacity of water.

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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. –  Bernhard Apr 26 '12 at 10:43
    
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? –  Pygmalion Apr 26 '12 at 11:13
    
For water at $100^\circ C$, do we have to take into account latent heat? –  leongz Apr 26 '12 at 11:46
    
@leongz Only if you use vapour instead of liquid water. –  Pygmalion Apr 26 '12 at 12:11
    
@Pygmalion: This only works due to the constant specific heat of water, which is a reasonable approximation. –  Ron Maimon Apr 28 '12 at 5:44
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