100 milliliters of 10 degrees celsius water plus 100 milliliters of 90 degrees celsius water, what is the final temperature?

Answer 1

simplified energy balance:
Q,absorbed + Q,lost = 0
where Q = mc(T2-T1)
in which Q=heat in J, m=mass in kg, c=specific heat capacity (which has a definite value for various substances) in J/(kg*C), T=temperature in Kelvin
*T2=final temp (the same for both and is usually the unknown)
*T1=initial temp

since in this problem, it involves only one type of substance (which is water), the c will be canceled:
Q,absorbed + Q,lost = 0
m1*c*(T2-T1,a) + m2*c*(T2-T1,b) = 0
m1*(T2-T1,a) + m2*(T2-T1,b) = 0
..from here, recall that density is mass per unit volume or:
d=m/V
density of water is approximately equal to 1 g/mL, thus,
m1 = 100 g
m2 = 100 g
since they are the same, we cancel them from the equation:
m1*(T2-T1) + m2*(T2-T1) = 0
(T2-T1,a) + (T2-T1,b) = 0
solving for T2:
T2 - 90 + T2 - 10 = 0
2(T2) = 100
T2 = 50 degrees Celsius

*note that change in T is involved thus Celsius and Kelvin unit can be used interchangeably

so there,, sorry for long explanation~

Answer 2

t2=50 degrees celisius

Answer 3

10 + 90 = 100
100 / 2 = 50
Example:
(-4)+(-6)20 + 12 + 8 + 24 + 9 = 63
63 / 7 = 9