100 milliliters of 10 degrees celsius water plus 100 milliliters of 90 degrees celsius water, what is the final temperature?
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~
10 + 90 = 100
100 / 2 = 50
Example:
(-4)+(-6)20 + 12 + 8 + 24 + 9 = 63
63 / 7 = 9
To find the final temperature, we can use the principle of heat transfer known as the Law of Mixtures.
The Law of Mixtures states that when two substances with different temperatures mix together, the final temperature is the average of their initial temperatures weighted by their respective masses or volumes.
In this case, we have 100 milliliters (ml) of water at 10 degrees Celsius and another 100 ml of water at 90 degrees Celsius.
First, let's calculate the initial heat content (Q) for each volume of water using the equation:
Q = mass (in grams) × specific heat capacity × temperature change
The specific heat capacity of water is about 4.18 Joules per gram Celsius (J/g°C).
For the first volume of water at 10 degrees Celsius:
Q1 = 100 ml × 1 g/ml × 4.18 J/g°C × (10°C - Tf)
For the second volume of water at 90 degrees Celsius:
Q2 = 100 ml × 1 g/ml × 4.18 J/g°C × (Tf - 90°C)
Since the total heat content remains constant after mixing, we can equate Q1 and Q2:
Q1 = Q2
100 ml × 1 g/ml × 4.18 J/g°C × (10°C - Tf) = 100 ml × 1 g/ml × 4.18 J/g°C × (Tf - 90°C)
Now, we can solve for Tf (the final temperature):
100 ml × (10°C - Tf) = 100 ml × (Tf - 90°C)
1000 ml°C - 100 ml×Tf = 100 ml×Tf - 9000 ml°C
200 ml×Tf = 1000 ml°C + 9000 ml°C
200 ml×Tf = 10000 ml°C
Tf = 10000 ml°C / 200 ml
Tf = 50°C
Thus, the final temperature after mixing 100 ml of 10°C water with 100 ml of 90°C water will be 50 degrees Celsius.