Physics/Chemistry problem
Author
A 150.0g sample of a metal at 75.0C is added to 150.0g of water at 15.0C. The temperature of the water rises to 18.3C. Calculate the specific heat capacity of the metal, assuming that all the heat lost by the metal is gained by the water.
Ooh, a double post! As can be seen in the Math and Physics forum...
I wonder if you realize that the purpose of these forums is not to be an "answer service," but rather a place for discussion, debate and learning. Getting someone else to do your homework isn''t learning.
And if you can''t figure that Qmetal = Qwater, then you truly are dumb.
Don''t worry about me; I''m the blunt one.
I wanna work for Microsoft!
I wonder if you realize that the purpose of these forums is not to be an "answer service," but rather a place for discussion, debate and learning. Getting someone else to do your homework isn''t learning.
And if you can''t figure that Qmetal = Qwater, then you truly are dumb.
Don''t worry about me; I''m the blunt one.
I wanna work for Microsoft!
Author
Dude, of course Heat(water) = Heat(metal)
specific heat of water = 4.18 J/Cg
delta T for water = 3.3 degrees so
Heat(water) = (4.18)(150)(3.3) = 2069 J
But I don''t know the temperature change of the metal.
specific heat of water = 4.18 J/Cg
delta T for water = 3.3 degrees so
Heat(water) = (4.18)(150)(3.3) = 2069 J
But I don''t know the temperature change of the metal.
Author
Of course, after a certain amount of time. The problem should''ve state the time or makes it clear.
quote:Original post by vbisme
But I don''t know the temperature change of the metal.
quote:Original post by vbisme
Calculate the specific heat capacity of the metal, assuming that all the heat lost by the metal is gained by the water
What else do you need?
I wanna work for Microsoft!
quote:Original post by vbisme
Of course, after a certain amount of time. The problem should''ve state the time or makes it clear.
Time is irrelevant to the solution of this problem. Go read your textbook.
I wanna work for Microsoft!
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