course phy 122 ??K????????l??y?assignment #012012. `query 2
......!!!!!!!!...................................
14:50:12 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
......!!!!!!!!...................................
RESPONSE --> the unknown specifi heat can be found by taking the temperature change of the water, the mass of the water, and multiplying the two to get x, x can then be multipled by 4180 joules, to find ho many joules are lost, in a closed system it can be assuemd that these joules were not lost, but used to raise the heat of the other item in the system. divide the joules by the temepratyre change and the weight, and you have the specific heat
.................................................
......!!!!!!!!...................................
14:50:17 ** 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. **
......!!!!!!!!...................................
RESPONSE --> k
.................................................
......!!!!!!!!...................................
14:53:19 prin phy Ch 13.26. Kelvin temperatures corresponding to 86 C, 78 F, -100 C, 5500 C and -459 F.
......!!!!!!!!...................................
RESPONSE --> you muct be kidding, 0 c is equal to 273 k so 86c= 259k -100c= 173k 5500c=5773k 78F=298.7k -459f=.4k
.................................................
......!!!!!!!!...................................
14:53:25 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.
......!!!!!!!!...................................
RESPONSE --> ya
.................................................
......!!!!!!!!...................................
16:13:21 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.
......!!!!!!!!...................................
RESPONSE --> pv/t=pv/t so 1*1/193= 1/9*40/x 1/9*40 is 4.44444444 1/193= .00518134715025906... final temp is 857.7 K
.................................................
......!!!!!!!!...................................
16:13:30 query gen phy ch 13.38 fraction of air released after tire temp increases from 15 to 38 C at 220 kPa gauge
......!!!!!!!!...................................
RESPONSE --> principles
.................................................
......!!!!!!!!...................................
16:13:35 ** 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. **
......!!!!!!!!...................................
RESPONSE --> k
.................................................
......!!!!!!!!...................................
16:13:40 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?
......!!!!!!!!...................................
RESPONSE --> k
.................................................
......!!!!!!!!...................................
16:13:44 ** 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
......!!!!!!!!...................................
RESPONSE --> k
.................................................
......!!!!!!!!...................................
16:13:48 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. **
......!!!!!!!!...................................
RESPONSE --> k
.................................................
......!!!!!!!!...................................
16:13:51 univ phy 17.115 time to melt 1.2 cm ice by solar radiation 600 w/m^2, 70% absorption, environment at 0 C.
......!!!!!!!!...................................
RESPONSE --> k
.................................................
......!!!!!!!!...................................
16:13:54 ** 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. **
......!!!!!!!!...................................
RESPONSE --> k
.................................................
"