Question

If you have 400.0 mL of water at 25.00 °C and add 140.0 mL of water at 95.00 °C, what is the final temperature of the mixture? Use 1.00 g/mL as the density of water.

Answer 1

400 mL water=400*1 g water=400 g water
similarly 140 mL water=140 g water
let the final temperature be \(\theta \)
so \[m _{1}s(\theta _{2}-\theta)=m _{2}s(\theta -\theta _{2})\]

Answer 2

here \(m _1\)=140 g \(\theta _2\)=95 degree C
and \(m _2\)=400 g \(\theta_1\)=25 degree C
s= specific heat of water