10 mL of 60oC water is mixed with 10 mL of 0oC water in a perfect calorimeter. What is the final temperature?
I hope that is not 600 C water, but 60C water. (under SI, no degree symbol is used).
equal masses? 30C
I understand that it is 30C, but what equation would one use for this?
If you want a formula for use with any amounts, then
[mass cool water x specific heat x (Tfinal-Tinitial)] + [mass warm water x specific heat x (Tfinal-Tinitial)] = 0
To find the final temperature, we can use the principle of conservation of energy, specifically the equation for heat transfer:
q1 + q2 = 0
where q1 represents the heat gained by the 60°C water, q2 represents the heat gained by the 0°C water, and the sum of these heats is zero because no energy is gained or lost within the perfect calorimeter.
Let's calculate the heat gained or lost by each water sample:
q1 = (mass1) × (specific heat capacity1) × (change in temperature1)
q1 = (10 g) × (4.18 J/g°C) × (final temperature - 60°C)
q2 = (mass2) × (specific heat capacity2) × (change in temperature2)
q2 = (10 g) × (4.18 J/g°C) × (final temperature - 0°C)
Since q1 + q2 = 0, we can set these equations equal to each other:
(10 g) × (4.18 J/g°C) × (final temperature - 60°C) = -(10 g) × (4.18 J/g°C) × (final temperature - 0°C)
Canceling out the common terms, we get:
final temperature - 60°C = -final temperature
Simplifying:
2 × final temperature = 60°C
Dividing both sides by 2, we find:
final temperature = 30°C
So, the final temperature after mixing the 60°C water with the 0°C water is 30°C.