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 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?
5 answers
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.
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)
the water cools down
25 * .897 * (40-0) = 100 * 4.186 * (Ti-40)
42.1 degrees celsius