- #1
mateomy
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I think I might have a hard time adequately explaining my issue, but here we go…
So everyone knows the Q=mc(delta)T equation, I have this problem that I am working on and it was driving me CRAZY! So I checked out the solutions manual after some time.
Basically the stated a mass of copper was sliding across a slab of ice (Ice, air, and copper were all at 0 degrees Celsius), it said the block eventually comes to rest due to friction. It only gave the mass of the copper (1.60kgs) and the speed (2.50 m/s) and wanted to know the mass of the ice that melted due to the sliding copper. It mentioned looking at the internal energy of the system; both ice and copper, and determining for each what the changes were.
Here's the solution to the first part of the problem…
The thing that I am not sure about is substituting the Kinetic value as the Q value as shown in the latent heat calculation. Can you do that? Can you just substitute any energy as the "Q" variable?
Does that make sense? I am on about 15 hours of studying and I think I might be brain-dead.
Thanks.
So everyone knows the Q=mc(delta)T equation, I have this problem that I am working on and it was driving me CRAZY! So I checked out the solutions manual after some time.
Basically the stated a mass of copper was sliding across a slab of ice (Ice, air, and copper were all at 0 degrees Celsius), it said the block eventually comes to rest due to friction. It only gave the mass of the copper (1.60kgs) and the speed (2.50 m/s) and wanted to know the mass of the ice that melted due to the sliding copper. It mentioned looking at the internal energy of the system; both ice and copper, and determining for each what the changes were.
Here's the solution to the first part of the problem…
The thing that I am not sure about is substituting the Kinetic value as the Q value as shown in the latent heat calculation. Can you do that? Can you just substitute any energy as the "Q" variable?
Does that make sense? I am on about 15 hours of studying and I think I might be brain-dead.
Thanks.