Question

What could be the final temperature of a
mixture of 100 g of water at 90°C and 600g
of water at 20°C.
C
(Ans. 30°C)​

Answer 1

Answer: mark as brainliest if it helps

We know that energy E = mCpΔT where m is the mass, Cp is the heat capacity (for water, 4.186J/g) and ΔT is the change of temperature.

In this case, E1 is the temperature decrease of the 100g of water at 90C

E1 = 100g x 4.186J/g-C x (90 – Tf)C where Tf is the final temperature

E2 = 600g x 4.186J/g-C x (Tf – 20)C

From E1 = -418.6Tf + 37674

From E2 = 2511.6Tf – 50232

But E1 = E2, so

-418.6Tf + 37674 = 2511.6Tf -50232

From which,

Tf = 30C

Explanation: