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Early Monday moring the temperature in the lecture hall has fallen to 398F, the same as the temperature outside At7.00 AM the janitor turns on the furace with theth...

Question

Early Monday moring the temperature in the lecture hall has fallen to 398F, the same as the temperature outside At7.00 AM the janitor turns on the furace with thethermostat set at 75'F The time constant for the building is K = 3hr and that for the building along with its heating system is hr Assuming that the outside temperature remains constant, what will be the temperature inside the ecture hall at 8.30 AM When will the temperature inside the hall reach 72*F?

Early Monday moring the temperature in the lecture hall has fallen to 398F, the same as the temperature outside At7.00 AM the janitor turns on the furace with the thermostat set at 75'F The time constant for the building is K = 3hr and that for the building along with its heating system is hr Assuming that the outside temperature remains constant, what will be the temperature inside the ecture hall at 8.30 AM When will the temperature inside the hall reach 72*F?



Answers

Early Monday morning, the temperature in the lecture hall has fallen to $$40^{\circ} \mathrm{F}$$, the same as the temperature outside. At 7:00 a.m., the janitor turns on the furnace with the thermostat set at $$70^{\circ} \mathrm{F}$$. The time constant for
the building is 1>K = 2 hr and that for the building
along with its heating system is 1>K1 = 1>2 hr. Assuming that the outside temperature remains constant, what will be the temperature inside the lecture hall at 8:00 a.m.? When will the temperature inside the hall reach $$65^{\circ} \mathrm{F}$$?

The base of the equation we will be using is TFT is equal to zero plus t sub zero minus and sub zero e to the power with native Katie. We know that absolute zero's. He put 12 piece of jurors equal to 21 K is equal to 1/3 looking in values for tea, sub zero, m sub zero and Kay. We end up with T if he's equal to 12 plus nine e to the power of negative one over three teeth. Since we want to find when the building will be 60 degrees, we set T is equal to 16 and then solve for tea. End up with T is approximately equal. Two negatives. Three times ln of four over nine hours.

In this problem. Temperature Capital T is given by the function f of X and T, where X is the distance away from a heater and time is the Ortiz the time after the heater is turned on, Um, we are told that at a time later of five minutes, so At T equals five and at a distance of x meters. I'm sorry at a distance of 3 m or X equals three, um, that the temperature changes at specific rates with respect to time and that with respect to um distance, we're told that the temperature increases at 1.2 degrees Celsius permanent, so that means the rate of change with respect to time, all other things held constant is given by 1.2 meters. I'm sorry, 1.2 degrees Celsius per minute. We're also told that the temperature decreases at two degrees per meter away from the um away from the heater. So that means that the derivative with respect to distance or X is given at negative two. Because, um, with increasing X values, the temperature decreases. So, um, we have to find a We have to find Delta T from this point t equals five and X equals three to appoint later where t equals six minutes and X equals 2.5 m away from the, um away from the heater. In order to do that, we need to approximate or we need to make the assumption that these derivative values here are going to be constant over that range. And that's an approximation we could make because the interval from 5 to 6 and the interval from 2.5 to 3 are small intervals, relatively speaking. So what we do is we say our change and, uh, our change in temperature is going to be our change of temperature. With respect to distance, times are change of distance plus the change of temperature with respect to time, times our change in time. Okay. And once again, this is something that we can. This is a calculation that we can on Lee make under the assumption that the derivatives do not the derivatives do not change over this interval. So we simply plug in the values that we have. Our Delta X is going to be three minus Alright, Sorry. 2.5 minus three. So that's negative in this case. And that makes sense because we're getting closer to the heater from 3 to 2.5. Our value is decreasing so that our Delta X has to be measured in negative. Our, um, derivative with respect to X was also negative. That was negative two. And then we plug in 1.2 degrees Celsius per minute and delta T is from six from five minutes to six minutes. Six minus five is one minute. Alright, so then we perform algebra and we get that. The change in temperature is 2.2 degrees Celsius. Very good.

In this problem. His student living in a room turns her 151. What fan on in the morning, We wish to calculate the temperature in the room when she comes back 10 hours later. We assume that is an ideal guess that the kinetic and potential energy changes is negligible. We can use constant specific heats at room temperature for a and all the doors and windows were tightly closed, so he transfer is disregarded. Now the gas constant on Air R is equal to zero point 287 You know, Paschal's Meet a Cube per K G. Calvin's our from our tables, and the specific heat for a room temperature is 0.718 killer jewels, but K G Calvin. Now we'll take the room as the system and the system is closed. Since the doors and windows are said to be tightly shut and no mess can into the system or leave the system so we can hands right. The energy balance equation for the system, as the energy in minus the energy out, is equal to the change in energy of the system data mhm the contributions off these energies is only due to the electrical energy from the fan. So we have the work done by defend. W e in is responsible for the changing internal energy in the room. Dr. U and Delta. You can written as m the mass off air into he's specific internal energies off state to minor state one which is in the morning. So we can write this as m times the specific heat see month by by a change in temperature T two in the afternoon. Man is t one in the morning. Now, to use this equation, we will firstly find the mass off the air. So the massive air we can use the relation p one times constant volume V off our T one and the volume V remember V is equal to three times for times four and that's 48 cubic meters. So we can substitute that into this equation. So the pressure is dead. Pressure 100. You know Paschal's The volume is 48 cubic meters, divided by the gas constant are off zero point 287 kilo Pascoal Meter cubed O k G Calvin and the temperature t one in the morning is 293 Calvin. So the mass of a we find is 57.8 k g. Now that we have the massive where we can go back, we can go back to. But before we go back to that equation to the energy balance equation, you know that the work done by the fan the electrical work done W e is equal to the rate at which the work is done. The power output off the electric fan multiplied by Dr T. So we are given the power output as 0.1 kilowatts or 0.1 kg jewels the second and a time off eight hours and each hour has 3600 seconds. And so we get the work done. Electrical work done by the fan to be 2000 800 and 80 kilo jokes. Yeah. Now we can go back to the energy balance equation and we confined t two. So the energy balance equation tells us that 2880 killer jewels the work done by the electric fan is equal to the mess off the air. 57 0.8 kg times three Specific heat off a room temperature 0.718 You know, ju for K G degree Celsius times T two minus 20 degree Celsius. So if you rearrange this equation, we confined the final temperature of the room in the evening, and T two is equal to 90 0.3 degrees Celsius. So we note here that the fan actually causes the internal temperature off the confined space to rise.

Using the derivative off tea with respect to t equals Mhm time in Yeah, minus C t. Let's got he. That's Hey, you times td minus t 40 This is Equation one with Qiyue equals key one quite. The ski, which equals T minus 0.5, equals 2.5 and mhm. He equals zero. Okay, Equals 0.5. Name off the equals 35 degrees and TD equals 16. The group, by substituting into one, we get the derivative the for the spectra T. It will 0.5 times 35 minus he less 2.5 times 16 minus t. Yeah. So the derivative he equals 17.5 minus 0.5. He most 40 minus 2.5. He which equals 57.5 minus three. He so derivative. 50. That's the times T equals 57.5. And this is a question, too. The soul, this equation, we obtain an integrating factor. We're be he equals three. So the integral bt DT equals three d t which equals three times t and so you he equals exponential off the integration p the okay equals financial off three t by multiplying the question too with you off? Um U of T. We get exponential. Well, three time the the Y T plus see potential off treaty time he see equals 57.5 times 20 show the so the derivative Well, Friendship 50. Thanks. T equals 57.5. Venture off the tee on By integrating both sides, we get both treaty times He equals integral Well, 57.5 He city DT equals 115. We'll work take times Ive well our 50. Yes. Constant See one so we can see tea party equals 115 over six less See, don't he? For negative T t who in the value off? See, we have you zero equals 55 So 55 equals 115 over six. Thus see, so we have seen equals 215 over tea. Therefore see he equals 150 15/6 plus 215 over six times e our negative three t finally to know all the temperature inside the van reached 27 degrees Thief off the equals as you see 100 of hell. Six plus 215 over six deep or negative the T we would that equal to 27 and we get that 47 or what? Six equals 215 over six comes he or negative. The T and 47 over 200 team equals he or negative three t By taking Berlin for both sides, then 47 over 215 he was negative three t and so he will approximately equal toe point 507 Therefore, the temperature inside the van would reach 27 degrees. At T equals 0.5 or seven hours or 15.42 men. Thank you.


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