Determining the precentage of mass in mixing two different temperature of water

How do I calculate the percentages to mix water at 1 degree to water at 250 degress to get 80 degress?

 

Answer:

you can use "black" principle:

mhot*c*(Thot-Tmix)=mcold*c*(Tmix-Tcold)

where:
mhot: mass of water 250 degrees
mcold: mass of water 1 degrees
Thot: 250 degrees
Tcold: 1 degrees
c: heat specific

because all of them is water, so they have same heat specific, then:

mhot*c*(250-80)=mcold*c*(80-1)

mhot/mcold = (250-80)/(80-1)=2.15

it means to get 80 degrees, the ratio of hot and cold water is 2.15.
if you have mhot 2.15kg, then you need 1 kg mcold

so the percentage of hot water: 2.15/(2.15+1)*100%=68.25%
and percentage of cold water : 1/(2.15+1)*100%=31.75%