#$&* course phy 242 june 13 around 2:40 am Question: query problem 15 introductory problem sets temperature and volume information find final temperature.
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Given Solution: ** PV = n R T so n R / P = V / T Since T and V remain constant, V / T remains constant. • Therefore n R / P remain constant. • Since R is constant it follows that n / P remains constant. ** STUDENT QUESTION: I don’t understand why P is in the denominator when nR was moved to the left side of the equation INSTRUCTOR RESPONSE: The given equation was obtained by dividing both sides by P and by T, then reversing the sides. We could equally well have divided both sides by v and by n R to obtain P / (n R) = T / V, and would have concluded that P / n is constant. To say that P / n is constant is equivalent to saying the n / P is constant. Your Self-Critique: ok Your Self-Critique Rating: ok ********************************************* Question: why, in terms of the ideal gas law, is T / V constant when only temperature and volume change? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: In terms of ideal gas law, the reason for T/V is constant is that because the other side of the equation is constant as well :P/nR. If T/V was not constant then there would be a proportionality where when one increased the other would increase as well to maintain the other side of the equation. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT ANSWER AND INSTRUCTOR RESPONSE: They are inversely proportional. They must change together to maintain that proportion. INSTRUCTOR RESPONSE: You haven't justified your answer in terms of the ideal gas law: PV = n R T so V / T = n R / P. If only T and V change, n and P don't change so n R / P is constant. Therefore V / T is constant, and so therefore is T / V. You need to be able to justify any of the less general relationships (Boyle's Law, Charles' Law, etc.) in terms of the general gas law, using a similar strategy. ** Your Self-Critique: ok Your Self-Critique Rating: ok ********************************************* Question: `q001. The temperature of a certain object increases from 50 Celsius to 150 Celsius. What is its change in temperature in Celsius? Convert 50 Celsius and 150 Celsius to Kelvin. What is the change in temperature in Kelvin? What is the change in temperature in Fahrenheit? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: There is +100 degrees Celsius change in temperature. 50 C= 323 K, 150 C = 423 K There is + 100 K change in temperature. There is +180 F change in temperature. confidence rating #$&*:out of 10 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-Critique Rating: OK ********************************************* Question: `q002. A sample of gas originally at 0 Celsius and pressure 1 atmosphere is compressed from volume 450 milliliters to volume 50 milliliters, in which state its pressure is 16 atmospheres. What is its temperature in the new state? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: 1) PV=nRT P=1 Atm V= 0.450 L N=? R= 0.082057L*atm/mol*K T=273 K 1 atm* 0.450 L = n *(0.082057L*atm/mol*K)* 273 K n= 0.02009 moles 2) PV=nRT P=16 Atm V= 0.050 L N=0.02009 moles R= 0.082057L*atm/mol*K T=? 16 atm* 0.050 L = 0.02009moles *(0.082057L*atm/mol*K)* T T= 485.3 K confidence rating #$&*:out of 10 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-Critique Rating: OK ********************************************* Question: `q003. What product or ratio involving P, V, n and T would remain constant if V and T were held constant? Why does this make sense? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-Critique Rating: ********************************************* Question: query univ phy 17.114 / 17.116 (15.106 10th edition) 1.5 * 10^11 m, 1.5 kW/m^2, sun rad 6.96 * 10^8 m. How did you calculate the total radiation of the Sun and how did you use this result to get the radiation per unit area? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: 1. Finding the surface area of the radiation it emits considering the Intensity : SA = 4*pi*r^2*I = 4*pi*(1.5*10^11m)^2*1500w/m^2 =4.2*10^26 W=H Power per unit must be attained: H/A = (4.2*10^26 W)/(4*pi*(6.96*10^8 m)^2)=6.97*10^7 W/m^2 Next, to find the temperature of the surface of the sun : H/A= σ*T^4 6.97*10^7 W/m^2 = (5.67*10^-8 W/m^2*k^4) *T^4 T= 5921 K confidence rating #$&*: 7 out of 10 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Outline of solution strategy: If we multiply the number of watts per unit of area by the surface area of the Sun we get the number of watts radiated from the Sun. The energy flows outward in a spherically symmetric manner; at any distance the entire power is distributed over the radius of a sphere concentric with the Sun and of radius equal to the distance. So if we divide that number of watts by the area of a sphere whose radius is equal to that of the Earth’s orbit, we get the number of watts per unit of area at that distance. This strategy is followed in the student solution given below: Good student solution: Surface area of sphere of radius r is 4 pi r^2; if flux intensity is I then flux = 4 pi r^2 I. When r = 1.5 * 10^11 m, I = 1500 W / m^2, so the flux is 4 pi r^2 I = 4 pi * (1.5 * 10^11 m)^2 * 1500 W / m^2 = 4.28 * 10^26 watts. 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, then T would be found as follows: 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. ** Your Self-Critique: ok Your Self-Critique Rating: ok ********************************************* Question: univ phy (omitted from 12th edition, but should be worked now) was 17.115 Solar radiation of intensity 600 watts / m^2 is incident on an ice sheet. The temperature above and below the ice sheet is 0 Celsius. Assuming that 70% of the radiation is absorbed at the surface of the ice, how long take to melt a layer of ice 1.2 cm thick? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: 70% of 600 watts/ m^2 : 70/100 = x/600 , x = 420 watts/ m^2 The units watts= joules/ second so therefore we can ma the unit transition to 420 joules/ second / m^2 Now we find the volume of the ice : 0.012 m^3 To find the mass of the ice we multiply the density of ice by the volume: 920 kg/m^3 *.012 m^3 = 11.04 kg. It takes 335,000 J to melt 1 kg of ice at 0 C, so 11.04 *335000 J= 3,698,400 J 3698400J/(420 J/s) = 8805.7 s = 146.8 minutes = 2.4 hours. confidence rating #$&*:6 out of 10 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** Thermal energy is not radiating in significant quantities from the ice, so only the incoming radiation needs to be considered, and as stated only 70% of that energy is absorbed by the ice.. • 70% of the incoming 600 watts/m^2 is 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. Thus each square meter absorbs 420 Joules of energy per second. We need to consider the volume of ice corresponding to a square meter. Having found that we can determine the energy required to melt the given thickness: • 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 obtain a more accurate result by using the a more accurate density; the density of ice (which floats in water) is actually somewhat less than that of water). • It takes about 330,000 Joules to melt a kg of ice at 0 C, so to melt 12 kg requires around 4,000,000 J. At 420 Joules/sec this will require roughly 10,000 seconds, or around 3 hours. All these calculations were done mentally and are therefore approximate. You should check them yourself, using appropriately precise values of the constants, etc. ** ********************************************* Question: `q004. A body with a 1/2 m^2 surface area, at temperature 25 Celsius, has an emissivity of approximately 1. It exchanges energy by radiation with a large surface at temperature -20 Celsius. At what net rate does it lose energy? By what percent would the rate of energy loss change if the large surface was at temperature -270 Celsius? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: P= A*e*sigma*T^4 Pout =.5 m^2*298 K ^4* 1*5.67*10^-8 W/m^2*k^4 = 223.6 W Pin= .5m^2*5.67*10^-8 W/m^2* k^4*253 K ^4 = 116.2 W Power out minus power in is the net power output (energy rate). 223.6 -116.2= 107.4 W Pout =.5 m^2*298 K ^4* 1*5.67*10^-8 W/m^2*k^4 = 223.6 W Pin= .5m^2*5.67*10^-8 W/m^2* k^4*(3 K) ^4 = 2.3*10^-6 W 223.6-(2.3*10^-6)= 223.6 Comparing it to 107.4 , it would change about 108.2 %. confidence rating #$&*:t of 10 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-Critique Rating:OK " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!