What would be the final temp of a mixture of 50gms of water at 20 degrees centigrade temp and 50gms of water at 40degrees centigrade temp
Use
T=(m1T1+m2T2)/(m1+m2)
=(50*20+50*40)/(50+50)
=30°C
Note: special case when the masses are equal, and the liquids are identical, the temperature is the mean of the two liquid temperatures.
To find the final temperature of a mixture of two substances, you can use the principle of conservation of energy. The total heat gained by the cooler substance will be equal to the total heat lost by the hotter substance until they reach thermal equilibrium.
The formula to calculate the final temperature of a mixture is:
(m1 * c1 * ΔT1) + (m2 * c2 * ΔT2) = 0
Where:
- m1 and m2 are the masses of the substances
- c1 and c2 are the specific heat capacities of the substances
- ΔT1 and ΔT2 are the changes in temperature of the substances
Let's apply this formula to the given problem:
In this case, we have 50 grams of water at 20 degrees Celsius (m1 = 50g, ΔT1 = Tf - 20°C), and 50 grams of water at 40 degrees Celsius (m2 = 50g, ΔT2 = Tf - 40°C).
The specific heat capacity (c) of water is approximately 4.18 J/g°C.
Substituting the values into the formula:
(50g * 4.18 J/g°C * (Tf - 20°C)) + (50g * 4.18 J/g°C * (Tf - 40°C)) = 0
Simplifying the equation:
(209 J/°C * Tf - 4180 J) + (209 J/°C * Tf - 8360 J) = 0
418 J/°C * Tf - 12540 J = 0
418 J/°C * Tf = 12540 J
Tf = 12540 J / 418 J/°C
Tf ≈ 30°C
So, the final temperature of the mixture would be approximately 30 degrees Celsius.