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Problem 1 Problem 2 Problem 3 Problem 4 Problem 5 Problem 6 Problem 7 Problem 8 Problem 9 Problem 10 Problem 11 Problem 12 Problem 13 Problem 14 Problem 15 Problem 16 Problem 17 Problem 18 Problem 19 Problem 20 Problem 21 Problem 22 Problem 23 Problem 24 Problem 25 Problem 26 Problem 27 Problem 28 Problem 29 Problem 30 Problem 31 Problem 32 Problem 33 Problem 34 Problem 35 Problem 36 Problem 37 Problem 38 Problem 39 Problem 40 Problem 41 Problem 42 Problem 43 Problem 44 Problem 45 Problem 46 Problem 47 Problem 48 Problem 49 Problem 50 Problem 51 Problem 52 Problem 53 Problem 54 Problem 55 Problem 56 Problem 57 Problem 58 Problem 59 Problem 60 Problem 61 Problem 62 Problem 63 Problem 64 Problem 65 Problem 66 Problem 67 Problem 68 Problem 69 Problem 70 Problem 71 Problem 72 Problem 73 Problem 74 Problem 75 Problem 76 Problem 77 Problem 78 Problem 79 Problem 80 Problem 81

Problem 50 Medium Difficulty

One winter day, the climate control system of a large university classroom building malfunctions. As a result, 500$m^{3}$ of excess cold air is brought in each minute. At what rate in kilowatts must heat transfer occur to warm this air by $10.0^{\circ} \mathrm{C}$(that is, to bring the air to room temperature)?

Answer

$\frac{Q}{t}=77.5 \mathrm{kW}$

Related Courses

Physics 101 Mechanics

College Physics for AP® Courses

Chapter 14

Heat and Heat Transfer Methods

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Thermal Properties of Matter

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Problem 14
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Problem 16
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Problem 18
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Problem 48
Problem 49
Problem 50
Problem 51
Problem 52
Problem 53
Problem 54
Problem 55
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Problem 58
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Problem 81

Video Transcript

in number 50. There's a malfunction in a heating system where 500 meters cubed of cold air is broken in each minute, and we're justifying what rate of heat transfers needed to raise the temperature by 10 degrees. It's a rate of heat transfer transfer, so that's, um, heat her tongue and we're not having to face change. So it's just changing temperature here. So that's a M C Delta T for the heat for time. So it's a rate. Um, we know of all you and we know that it's air and density of anything is its mass per volume so mass would be density times the volume. So I'm just gonna substitute that in here. So the rate of heat transfer it's gonna be density times, volume time, specifically times the change in temperature per time. I look on table 11.1 and I got the density of air was 1.29 and that was kilograms for meter, cubed I have my volume is in meters cubed specific heat. I looked up on a table for air table 14.1 and for air 721 and the change in temperature want is 10 degrees and that cold air was brought in 500 per minute. So, of course, my time has to be in seconds. I'm gonna make that 60. I multiply and, like, 77,500 And that would be wants. I'm asked to give him my answer and Kilowatts, though. So that would be 77.5 kilowatts.

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