When 200g of H2O at 90 C mix whit 20 g H2O at 0 C what is the final temp?

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gsenviro's profile pic

gsenviro | College Teacher | (Level 1) Educator Emeritus

Posted on

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.

electreto05's profile pic

electreto05 | College Teacher | (Level 1) Assistant Educator

Posted on

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

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