What would be the final temperature of a mixture of 50 g of 15 degrees C water and 50 g of 40 degrees C water?
Since there are equal amounts of water, the final temperature would be the mean of 15 and 40 C, which is 27.5 C.
To find the final temperature of a mixture of two substances, we can use the principle of heat transfer, which states that the heat gained by one substance equals the heat lost by the other substance.
The equation we can use is:
(m1 * c1 * ΔT1) + (m2 * c2 * ΔT2) = 0
where:
m1 is the mass of the first substance (water in this case),
c1 is the specific heat capacity of the first substance (water),
ΔT1 is the change in temperature of the first substance,
m2 is the mass of the second substance (water),
c2 is the specific heat capacity of the second substance (water),
ΔT2 is the change in temperature of the second substance.
In this case, the mass of both waters is 50 g, and the specific heat capacity of water is approximately 4.18 J/g°C.
Let's plug in the values:
(50 g * 4.18 J/g°C * ΔT1) + (50 g * 4.18 J/g°C * ΔT2) = 0
Now, let's rearrange the equation to solve for the final temperature (ΔTf):
ΔTf = -((ΔT1 * m1 * c1) + (ΔT2 * m2 * c2)) / (m1 * c1 + m2 * c2)
Using the given temperatures and substituting the values:
ΔTf = -((Tf - 15°C) * 50 g * 4.18 J/g°C + (Tf - 40°C) * 50 g * 4.18 J/g°C) / (50 g * 4.18 J/g°C + 50 g * 4.18 J/g°C)
Now, we can solve this equation to find the final temperature (ΔTf) of the mixture of the two waters.