i don't know what to do.
25 g of aluminum is added to 100 g of water. The aluminum is initially 0 degrees Celsius and its specific heat is 0.897 J/g*C. If the final temperature of both is 40 degrees Celsius, what was the initial temperature of the water?
heat lost by aluminum + heat gained by water = 0
heat lost by Al = mass Al x specific heat Al x (Tfinal-Tinitial)
heat gained by water = mass H2O x specific heat H2O x (Tfinal-Tinitial)
Substitute into the heat lost + heat gained = 0 and solve for Ti.
i don't get it? heat lost? ahhhh. this is very fustrating.
I don't like to do it this way because students ALWAYS get confusted so I'll put it together for you.
heat lost by Al is
mass Al x specific heat Al x (Tfinal-Tintial)
heat gained by H2O is
mass H2O x specific heat H2O x (Tfinal-Tinitial)
So set heat lost + heat gained = 0 and here it is.
[mass Al x specific heat Al x (Tfinal-Tinitial)] + [mass H2O x specific heat H2O x (Tfinal-Tinitial)] = 0
Now substitute the numbers and solve for Ti.
the Al heats up
the water cools down
25 * .897 * (40-0) = 100 * 4.186 * (Ti-40)
42.1 degrees celsius
To find the initial temperature of the water, we can use the principle of conservation of energy.
The equation used in this calculation is:
(mass of the substance) x (specific heat) x (change in temperature) = heat gained or lost.
First, let's calculate the heat gained or lost by the aluminum:
Heat gained or lost by aluminum = (mass of aluminum) x (specific heat of aluminum) x (final temperature - initial temperature of aluminum)
Heat gained or lost by aluminum = 25 g x 0.897 J/g*C x (40°C - initial temperature of aluminum)
Next, let's calculate the heat gained or lost by the water:
Heat gained or lost by water = (mass of water) x (specific heat of water) x (final temperature - initial temperature of water)
Heat gained or lost by water = 100 g x (4.18 J/g*C) x (final temperature of water - initial temperature of water)
Since the heat gained by the aluminum is equal to the heat gained by the water (according to the principle of conservation of energy):
(25 g x 0.897 J/g*C x (40°C - initial temperature of aluminum)) = (100 g x 4.18 J/g*C x (final temperature of water - initial temperature of water))
Now, we can solve for the initial temperature of the water:
(25 g x 0.897 J/g*C x (40°C - initial temperature of aluminum)) / (100 g x 4.18 J/g*C) = (final temperature of water - initial temperature of water)
Simplifying:
(25 x 0.897 x (40 - initial temperature of aluminum)) / (100 x 4.18) = (final temperature of water - initial temperature of water)
Now, plug in the given values:
(25 x 0.897 x (40 - 0)) / (100 x 4.18) = (40 - initial temperature of water)
Simplifying further, we get:
10.0585 ≈ 40 - initial temperature of water
Now, isolate the initial temperature of water:
initial temperature of water = 40 - 10.0585
initial temperature of water ≈ 29.9415°C
Therefore, the initial temperature of the water was approximately 29.9415°C.