00:01
So we have a lead ball, which i have a right as mass m sub l of 2 .3 kilograms, is placed in 2 .3 liters of water.
00:09
Well, one liter of water is about one kilogram of water, so that's a mass of the water, m sub w, of 2 .5 kilograms.
00:16
The initial temperature of the water, which i write as t sub i, w, so initial temperature of water is 20 degrees celsius.
00:23
And the final temperature of both the lead and the water in combination, or the equilibrium temperature, t, sub eq, is 32 degrees celsius.
00:30
And it wants us to find the initial temperature of the lead call, given this information.
00:34
We also have the constant values for lead and water, c -s -bel, c -s -w written out here as well, and units of joules per kilogram times degrees celsius.
00:46
Okay.
00:50
So the heat lost by the lead must be equal to the heat gained by the water in this situation.
00:57
So we're going to go ahead and write that out here.
00:59
We have m sub l times c sub l multiplied by the change in temperature, which was the initial temperature of the lead, t .i .l minus the new temperature, the equilibrium temperature...