Question

200 g of water at 90 degree celsius is mixed with the 100 g of alcohol at 10 degree celsius. Find the final temperature of the mixture.(specific heat of alcohol = 0.60 cal/g0c) *

Answer 1

Answer:

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

Strictly speaking, this is just an approximation, since Cp = Cp(T), so really,

E = m x (Integral of Cp(T) from T1 to T2)