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Hey everyone, this is question number 50 from chapter 14.
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In this problem, we're talking about a slab of material.
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We're given a cross -sectional area, a thickness, thermal conductivity, and then we're told that the temperature difference between the two faces is 80 degrees celsius, and we're asked to find how much heat flows through in one day.
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So we're given a conductivity constant.
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We're given thickness, cross -sectional area.
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So these should make you think of our conductivity equation.
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H equals k -a, conductivity area, th minus t -c over l, temperature gradient.
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Okay, so, oh, and then times t.
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And we know that this is just the change in q over, it's a relationship between q and t.
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So we can use q instead of h.
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It's just a constant q or just a letter to hold its place.
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But anyway, we have all this information now, how much heat flows through one day.
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So we're given conductivity 0 .075 watts per meters k...