00:01
Good day.
00:02
Heat is a form of energy that flows from a hotter to a forward body.
00:06
When the substance undergoes temperature change, then that heat associated is solved as q equals mc delta t, where q is the heat, then is the mass, c is the specific heat.
00:16
Delta t is the change in temperature solved as the difference between the final temperature tf and the initial temperature t .f.
00:22
When the substance undergoes phase change, which is from solid to liquid, liquid to gas, and so on, then the heat that is associated with that is solved as m times l, where m is the mass and l is the latent heat, which value depends on the type of phase change that of course.
00:39
In this problem, we want to know how much heat is released when water freezes, particularly when 8 .92 kilograms of water freezes at its freezing temperature that is 0 degrees celsius.
00:52
So the mass of 8 .92 kilograms and the freezing at the freezing temperature of zero degrees celsius so that's freezing we want to find the heat that is released for this case and then for letter b if the heat that is released from freezing water is absorbed by at 134 kilogram three we want to know by how much the temperature of of the tree would increase so delta t for that is going to be solved assume that the specific heat of the three is 2 .5 times 10 to the power of 3 joules per kilogram degrees celsius so let us start for letter a since the substance undergoes and particularly water undergoes space range then we will solve for so since freezing is involved, then we will use the latent heat of fusion of water, which is 334 joules per gram.
02:06
So the mass being 8 .92.
02:11
So let's first define that value that i said while go...