course Phy 202 I am a little behind this week, because my lap top crashed on Monday and I did not get it back until Wednesday, but I will try to have everything done by Sunday night. I just wanted to let you know, because I do not want you to think I am a slacker. Haha!
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11:39:18 query intro set problem 14 substance, water, both temperatures and masses known, final temperature known, find spec ht Explain how the final temperature of the system is combined with the known initial temperatures and masses of both substances to obtain the unknown specific heat
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RESPONSE --> 'dQ2 + 'dQ1 = 0 And: 'dQ2 = c2mc'dT2 'dQ1 = c1m1'dT1 So: [c2m2(Tf-T2)] + [c1m1(Tf -T1)] = 0 You can add 'dQ1 and 'dQ2 together to get zero because we assume a closed system with no energy gained or lost from the system and that the thermal energy lost by the water (m2) is gained by the substance (m1). In the problem, we are given m1, m2, Tf, T1, and T2, but we are not given either of the specific heats nor are we given the change in thermal energy. The question asked for the specific heat of the substance, which means I should probably be able to figure out the specific heat of the water from what I know and use that to help figure out the net change in thermal energy. Then I should be able to plug them into the equation to find the specific heat of the substance by solving the equation for c1. So, I looked up specific heat of water in the text and found out that it was 4186 J/kg*C. Now we can figure out the first part of the formula. You can set this equal to zero because the m1 * c1 * (Tf - T1) + m2 * c2 * (Tf - T2)=0 (8kg)c1(77.29333C-11C)+(12kg)(4186J/kg*C)(77.2933-99C) = 0 (8kg)c1(66.29333C)+(-1090369 J)=0 (8kg) c1 (66.29333C) = 1090369J c1 = 1090369J/(8kg * 66.29333C) c1 = 2055.96 J/kg*C which is the specific heat of the substance
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11:40:24 ** The change in the thermal energy of an isolated system is 0. So assuming that the systems are isolated the thermal energy change of one object is equal and opposite to that of the other. For an ideal substance the change in the thermal energy of an object is proportional to the change in its temperature and to its mass (i.e., more mass and more change in temperature implies more change in thermal energy). The specific heat is the proportionality constant for the substance. Using `dQ for the change in the thermal energy of an object we can express this as `dQ = mass * specific heat * `dT. (General College and University Physics students note that specific heat is not in fact strictly constant and for most substances changes with temperature.) We know that what energy one substance loses, the other gains, and we know the specific heat of water , so we can set the two quantities equal and solve for specific heat of the unknown substance. For two objects combined in a closed system we have `dQ1 + `dQ2 = 0, which gives us the equation m1 c1 `dT1 + m2 c2 `dT2 = 0 or equivalently m1 c1 `dT1 = - m2 c2 `dT2. If we know all the quantities except, say, the specific heat c2 of the second substance we can easily solve for c2. **
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RESPONSE --> Okay, sorry my explanation was so long. But I wanted to work out the whole problem for extra emphasis.
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11:42:05 prin phy Ch 13.26. Kelvin temperatures corresponding to 86 C, 78 F, -100 C, 5500 C and -459 F.
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RESPONSE --> Not a principles of physics student.
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11:42:12 The Kelvin temperature is 273 above the Celsius temperature (actually 273.15 below, but the degree of precision in the given temperatures is not sufficient to merit consideration of tenths or hundredths of a degree). 86 C, -100 C and 5500 C are therefore equivalent to ( 86 + 273 ) K = 359 K, -(100 + 273) K = 373 K, (5500 + 273) K = 5773 K. 78 F is (78 F - 32 F) = 46 F above the freezing point of water. 46 Fahrenheit degrees is the same as (5/9 C / F ) * 46 F = 26 C above freezing, or since freezing is at 0 C, just 26 C. The Kelvin temperature is therefore (26 + 273) K = 299 K. -459 F is (459 + 32) F = 491 F below freezing, or (5/9 C / F) * (-491 F) = 273 C below freezing, which is -273 C or (-273 + 273) K = 0 K. This is absolute zero, to the nearest degree.
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RESPONSE --> Okay
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12:09:08 prin phy and gen phy Ch 13.30 air at 20 C is compressed to 1/9 of its original volume. Estimate the temperature of the compressed air assuming the pressure reaches 40 atm.
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RESPONSE --> I tried to answer this question by using [(P1 * V1)/T1] = [(P2*V1)/T2], because we can assume that the number of moles of the air remain constant. So, solving for T2, the formula becomes: T2 = (P2 * V2) * [T1 * (P1 * V1)] V2 = (1/9)V1 So, T2 = (P2 * (1/9)V1) * [T1* (P1 *V1)] T2 = (40atm *(1/9)V1) * [293.15 K * (1atm *V1)] Now, all I need to do if figure out how to get V1 and I will be able to solve for T2, but I am not exactly sure how to do that. Maybe the V1's cancel out and then that would give you: T2 = (40atm *(1/9)) * [293.15 K * (1atm)] T2 = 1302.89K
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12:21:25 query gen phy ch 13.38 fraction of air released after tire temp increases from 15 to 38 C at 220 kPa gauge
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RESPONSE --> The first thing I did was to change C to K. 15C + 273 = 288K 38 C +273 = 311 K I used the formula (P1/V1) = (P2/V2) and solved for P2. 220kPa/288K = P2/311K (.764 kPa/K) = P2/311K P2 = (.764 kPa/K) * 311K P2 = 237.57 kPa Then I took the initial 220 kPa and divided it by 237.57 kPa to get the percentage of air that would remain in the tire and then subtracted that from 100% of the total air to get the amount that would have to be removed in order to keep the initial pressure in the tire. 220 kPa / 237.57 kPa = .926 = 93% 100% - 93 % = 7% or 7/100
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12:28:46 ** T1 = 288 K, T2 = 311 K so T2 / T1 = 311 / 288 = 1.08, approx. This is approx. an 8% increase in temperature. The pressure will end up at 3ll / 288 * 321 kPa = 346 kPa, approx (note that we have to use actual rather than gauge pressure so init pressure is 220 kPa + 101 kPa = 321 kPa, approx. You then have to change the number n of moles of gas to get back to 331 kPa, so n3 / n2 = P3 / P2 = 321 kPa / 346 kPa or approximately .93, which would be about a 7% decrease. Note that the results here are mental estimates, which might not be particularly accurate. Work out the process to see how the accurate numbers work out. **
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RESPONSE --> Okay.
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12:29:02 query univ phy 17.116 (15.106 10th edition) 1.5 * 10^11 m, 1.5 kW/m^2, sun rad 6.96 * 10^8 m. How did you calculate the total radiation of the Sun and how did you use this result to get the radiation per unit area?
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RESPONSE --> Did not have to do for General Physics.
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12:29:08 ** GOOD STUDENT SOLUTION The total radiation of the sun was the rate it reaches earth times the imaginary surface of the sphere from the sun center to earth atmosphere, or 1500 W/m^2 * (4`pir^2) = 1500W/m^2 * 2.8537 x10^23 m^2 = 4.28055 x 10 ^ 26 W. } Radiation per unit of area surface of the sun would be
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RESPONSE -->
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12:29:25 4.28055 x 10 ^ 26 W / (4*`pi * (6.96 x 10 ^ 8 m)^2) = 4.28055 x 10 ^ 26 W / 6.08735 x 10 ^ 18 m^2 = 70318775.82 J/s/m^2 = 7.03 x 10 ^ 7 J/s/m^2 If the sun is radiating as an ideal blackbody, e = 1, the T would be found: H = `dQ/`dt = 4.28055 x 10 ^ 26 W = (4*`pi * (6.96 x 10 ^ 8 m)^2) * (1) * (5.67051 x 10^-8 W/m^2*K) * T^4 So T^ 4 = 4.28055 x 10 ^ 26 W / 6.087351 x 10 ^ 18 m^2) * 1 * (5.67051 x 10^-8 W/m^2*K) T^4 = 1.240 * 10 ^ 15 K ^4 T = 5934.10766 K on surface of sun. **
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RESPONSE -->
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12:29:39 univ phy 17.115 time to melt 1.2 cm ice by solar radiation 600 w/m^2, 70% absorption, environment at 0 C.
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RESPONSE --> Did not have to do for General Physics
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12:29:43 ** Thermal energy is not radiating in significant quantities from the ice, so you use 70% of the incoming 600 watts/m^2, which gives you 420 watts / m^2, or 420 Joules/second for every square meter if ice. Melting takes place at 0 C so there is no thermal exchange with the environment. A 1.2 cm thickness of ice will have a volume of .012 m^3 for every square meter of surface area; the mass will be close to 1000 kg/m^3, so there are about 12 kg of ice for every m^2 of surface (you can correct this by using the correct density of ice). It takes about 330,000 Joules to melt a kg of ice, so to melt 12 kg requires around 4,000,000 J. At 420 Joules/sec the time required will be about 10,000 seconds, or around 3 hours. All my calculations are approximate and done mentally so you should check them yourself, using more precise values of the constants, etc. **
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