15 mL of 60oC water is mixed with 10 mL of 0oC water in a perfect calorimeter. What is the final temperature?
heat lost by hot water + heat gained by cold water = 0
[mass hot water x specific x (Tfinal-Tinitial)] +[mass cold water x specific heat x (Tfinal-Tinitial)] = 0
Solve for Tfinal.
-58,5
To find the final temperature after mixing the two water samples, we can apply the principle of conservation of energy.
The formula for heat transfer is:
Q = m * c * ΔT
Where:
Q = Heat transfer (in Joules)
m = Mass of the substance (in grams)
c = Specific heat capacity (in J/g°C)
ΔT = Change in temperature (in °C)
For the 15 mL of 60°C water, we need to calculate the mass using the density of water, which is 1 g/mL:
Mass = Volume * Density
Mass = 15 mL * 1 g/mL
Mass = 15 g
Now, let's calculate the heat transferred for the 15 mL of 60°C water:
Q1 = m * c * ΔT
Q1 = 15 g * c * (final temperature - 60°C)
For the 10 mL of 0°C water, we can immediately calculate the heat transferred:
Q2 = m * c * ΔT
Q2 = 10 g * c * (final temperature - 0°C)
Since the calorimeter is perfect, meaning there is no heat loss to the surroundings, the total heat transferred, Q1 + Q2, is equal to zero.
Q1 + Q2 = 0
Therefore:
15 g * c * (final temperature - 60°C) + 10 g * c * (final temperature - 0°C) = 0
Now, we can solve for the final temperature:
15 g * c * final temperature - 900 g°C + 10 g * c * final temperature = 0
25 g * c * final temperature = 900 g°C
final temperature = 900 g°C / (25 g * c)
The specific heat capacity of water is approximately 4.18 J/g°C, so:
final temperature = 900 g°C / (25 g * 4.18 J/g°C) ≈ 8.6°C
Therefore, the final temperature after mixing the two water samples is approximately 8.6°C.