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.