course Phy 202 ْސSIڴɸJassignment #012
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21:45:46 `questionNumber 120000 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 --> The initial temperature is subtracted from the final temperature to give you the change in temperature which is multiplied by the mass and the known specific heat. This energy is then divided by the second mass. Then that value divided by the change in temperature of the second object.
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21:49:12 `questionNumber 120000 ** 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 --> I made the mistake of assuming we knew the specific heat of one of the two objects. Instead I should have assumed that since it is an isolated system that the overal change in thermal energy is 0. `dQ= mass*specific heat*`dT
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21:49:14 `questionNumber 120000 ** 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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21:51:48 `questionNumber 120000 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 assigned
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21:52:07 `questionNumber 120000 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 --> ok
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22:03:08 `questionNumber 120000 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 --> ((P1V1)/T1) = ((P2V2)/T2) ((1 atm*1m^3)/20C) = ((40 atm (1/9 m^3))/T) ((1 atm*m^3)/20C) = (4.44 atm m^3)/T T ((1 atm*m^3)/20C) = (4.44 atm m^3) T (1atm*m^3) = 88.89 atm m^3 C T = 88.89 C
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22:03:10 `questionNumber 120000 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 -->
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22:06:25 `questionNumber 120000 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 --> PV/T = PV/T 220 kPa 1V/15 C = 220 kPa xV/38C 14.67 = 5.789 xV 2.53 = x = must change by this factor
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22:11:23 `questionNumber 120000 ** 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 --> My first mistake was not remembering to change from Celsius to Kelvin. I should have just found the percentage change in temperature and used it to find the pressure. Then I should have used the moles to get the ratio of pressure to find the percentage of pressure decrease.
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22:11:37 `questionNumber 120000 ** 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 --> not assigned
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22:11:40 `questionNumber 120000 ** 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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22:11:42 `questionNumber 120000 ** 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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22:11:45 `questionNumber 120000 ** 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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22:11:47 `questionNumber 120000 ** 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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22:12:19 `questionNumber 120000 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 --> ok
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22:12:21 `questionNumber 120000 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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22:12:24 `questionNumber 120000 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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22:12:26 `questionNumber 120000 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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