Question

If you combine 350.0 mL of water at 25.00 °C and 130.0 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

Ah, this is a thermochemistry question.
So we're basically using enthalpy for this--heat, I guess:

q = -q

The energy lost by the hotter water = energy gained by cooler water

\[q=-q\]\[mc \Delta(T)=-mc \Delta(T)\]
Plug in values and you're good to go.

Your knowns are:

m1=350.0 g (since multiplying the the density which is 1 over 1 will convert mL -> g)
c=4.184 J/goC
Ti1=25.00 oC

m2=130.0 g
c=4.184 J/goC
Ti2=95.00 oC

Plug in and solve for Tf, which will be the same for both because the mixture will reach thermal equilibrium.

\[-mc(Tf-Ti)=mc(Tf-Ti)\]

Answer 2

Thanks! :) @KinzaN

Answer 3

You're welcome ^^