If you combine 300.0 mL of water at 25.00°C and 110.0 mL of water at 95.00°C, what is the final temperature of the mixture?

1 Answer
Dec 23, 2016

Answer:

I got #44^@C# approx.

Explanation:

We can say that the heat lost by the hot water #Q_1# is equal to the heat #Q_2# gained by the cold where the heat can be written as funcion of mass #m#, specific heat #c# and the variation of temperature between final and initial temperature #DeltaT=(T_f-T_i)#.
We will have:
#Q_1+Q_2=0#
#m_1cancel(c)(T_f-T_1)+m_2cancel(c)(T_f-T_2)=0#

With #1ml=1mg# for mass:

#110(T_f-95)+300(T_f-25)=0#
#110T_f-10,450+300T_f-7,500=0#
#410T_f=17,950#
So:
#T_f=(17,950)/410=43.78~~44^@C#