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.
Best Answer
Heat transfers from hotter water to cooler water until temperatures equalize.
If mass and temperature of the hotter water are $m_H$ and $T_H$, 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
$$T = \frac{m_H T_H + m_C T_C}{m_C+m_H}$$