When 200g of H2O at 90 C mix whit 20 g H2O at 0 C what is the final temp?
When two quantities of water are mixed at different temperatures, a heat transfer occurs during the process. In this case the mass of colder water, absorbs a quantity of heat equal to that released by the hot water.
The equation of the amount of heat absorbed or released by a substance, for a variation in temperature is:
Q = m Ce (Tf – Ti), where:
Q: is the amount of heat absorbed or released
Ce: is the specific heat of the substance
m: is the mass of the substance
Ti: is the initial temperature T
Tf: is the final temperature T
Then we can write for each portion of water, the following equations:
For hot water:
- Qh = mh Ce (Tf – Th); the negative sign represents that heat is released.
For cold water:
Qc = mc Ce (Tf – Tc)
Equating the above equations:
- Qh = Qc
- (mh Ce (Tf – Th)) = mc Ce (Tf – Tc)
Now we solve this equation for the value of the final temperature Tf
- ((200 g) Ce (Tf – 90°)) = (20 g) Ce (Tf – 0°)
- 200 Tf + 18000 = 20 Tf
Tf = 81.8°C
So that, the final temperature of the mixture is 81.8 °C
A simpler method would be to use weighted averages.
Final temperature = (V1T1 + V2T2)/(V1 + V2)
= (200 x 90 + 20 x 0)/(200 + 20) = 18000/220 = 81.818 degree C
Hope this helps.