Given emissivity and area, is there a way to calculate how fast a blackbody would cool?
929275
You would need the specific heat of the material.  The rate of cooling would be the total energy loss (E*A) divided by the specific heat.  This will be non-linear, and your value would be good for only the specific temperature given.\n\nWatts = Energy/time; one watt = 1 joule/sec.
1128546
specific heat Q = mc(t2-t1)\nput everything in SI units..\nthen \n\n40,000= 500/1000 * c(68-20)\n40000=0.5* 48*c \nhence c= 40000/24= 1666.67 joules/kg oC.
501191
There is a formula...but this is much easier\n\nhttp://www.onlineconversion.com/temperature.htm
791308
it depends on the material it is being trasferd in ... and i guess the means of transfer.  \n\nI mean i can transfer heat via infared radiation ... and that is a light wave, so it transfers at the speed of light.  \n\nOn the other hand there is a propertie of substances that is differant called the "thermal conductivity", the matireals ability to conduct heat.  its units are joules per meter per kelvin per second.\n\nso it is the ammount of energy that it can transfer over a distance of one meter per degree (C) of differance in one second.  \n\nA typical value for metal is around 100, and for wood about 0.2.  \n\n(ok time for an example)\n\nlet us say that you put a rod of matiral in boiling water.  IT is one meter long, how long would it take to heat another pot of water.  \n\nSo it tranfers X joules of enengy every second over that one meter distance.  \nLets say you have 100 grams of watter on the other side and you want to see it rase in temp one degree ... that would take about 400 joules (you have too look that up) \n\nit would take 400/x seconds for a noticable ammount of heat to transfer a distance of one meter.  \n\nSo for metal about 4 four seconds, for wood about 1 hr.  \n\nFor dimond (the most theramly conductive substance on the planet with a conductivity of over 1000) it would take about a half a second\n\n(this is all very elementry and it does not take in to account a lot of things, but is is a good idea of how fast these things happen... but in reality you also have to heat up the matireal in order to feel the effect and all that stuff, but i ignore that ... i belive thta you just want the general idea of the speeds ...)\n\nSo for metal i would say some were around a meter per second, for wood I would say about 0.01 meters persecond.  \n\nFor air... well that is the speed of the wind that day.
1160653
Space near the earth is actually filled with a gas called the interplanetary medium (see first source). This gas, which is mostly gas coming from the sun, is quite hot, around 100,000 degrees C. But, it's also only 10^-18th the density of sea level air, so it carries very little heat.\n\nAir is relatively thick, so we're used to conduction and convection determining an objects temperature.  There's very little of either in space, so radiation dominates.  The sun's light is about twice as strong in space as it is on earth.  How hot this heats an object depends on the objects shape and color, but it's fair to say that most things will heat up to a few hundred degrees. \n\nBecause there's no air in space, everything looses heat very slowly, like being in a thermos. So, anything that produces heat, like a human body or machinery, will gradually warm up--space suits have built-in coolers to deal with this. \n\nIf you're out of the sunlight and have used up all your internal energy, you eventually cool down to about -25 to +25 C.  Infrared radiation from the earth is what prevents you from getting colder.  If you were also far away from the earth, you'd cool down to something between -240C and -270C.
397497
I would use the ideal gas law...more specifically...the combined ideal gas law...\n\nP1*V1/T1 = P2*V2/T2
742342
29.444..°C\n\nC = (F - 32) * (5/9)\nC = (85 - 32) * (5/9)\nC = 53 * 5/9\nC = 265 / 9\nC = 29 4/9° C
187220
W=F*d,  d=W/F,  d=600J/400N = 1.5m
715037
Ok-i'm not quite sure if i did this right--we covered this last year--so you might want to double check with your textbook or teacher, or whatever resource you have...\n\nwhat i would do is find the final temperature:\nFinal Temperature=(V2 * T1) / V1\n\nand then use the delta T, delta v, and P to find delta E.\n\nbut i'm really not that sure.\n\nsorry that i wasnt a better help!
604131
Extrapolate the formula, using the surface temoerature, and instead of the earth distance, substitute the distance you want to caluclate the temperature at.  \n\n\nMeasuring the Sun's Temperature\nOne method, called Wien's law, uses the wavelength of the peak emission, wavelengthpeak, in the Sun's continuous spectrum. The temperature in Kelvin = 2.9 × 106 nanometers/wavelengthpeak.\n\nAnother method uses the flux of energy reaching the Earth and the inverse square law. Recall from the Stellar Properties chapter that the flux is the amount of energy passing through a unit area (e.g., 1 meter2) every second. From the Inverse Square Law of Light Brightness, you find that the solar flux at the Earth's distance = the Sun's surface flux × (Sun's radius/Earth's distance)2 = 1380 Watts/meter2. Since the Sun's photosphere is approximately a thermal radiator, the flux of energy at its surface = sigma × (the Sun's surface temperature)4, where sigma is the Stefan-Boltzmann constant. Rearranging the equation, the photosphere's temperature = [(solar flux at Earth)/sigma) × (Earth distance/Sun's radius)2]1/4.\n\nThese two methods give a rough temperature for the Sun of about 5800 K. The upper layers of the photosphere are cooler and less dense than the deeper layters, so you see absorption lines in the solar spectrum. Which element absorption lines are present and their strength depends sensitively on the temperature. You can use the absorption line strengths as an accurate temperature probe to measure a temperature of about 5840 K. \n\nRead the link, you'll have to scroll down to come up to the stuff I have cut and paste above.\nst\nHope you find this useful.
791970
Use the Gay-Lussac gas law: V/T is constant. Note that T here is the temperature in Kelvin, not in Celsius. Luckily 27 C = 300 K
1119466
I donno ;=)\n\nMay be start with a heat balance equation like\n\nQfinal=Qwater-start  - Qice-start – Qlatent\n\nIn other words the final heat of the system is equal to the original amount of heat minus the heat required to bring ice to the melting point and minus again the amount of heat needed to melt the ice\n\nOr \nQfinal=(Specific-heat of water) (temp in K)(mass-final)\n\nmass-final= mass of water + mass of ice= m1 + m2\n\nNote -273.16 °C =0 degrees on Celsius scale = 273.16 °K (273 I hope is good enough for you?)\n\nQwater-start  =( 4190 J/(kg*K) (t-start K degrees )(m1)( 4190 J/(kg*K)=A(m1)  A= 4190 (76.2 + 273)\n\nQice-start= 2100 J/(kg*K)( - 23.6+273)(m2)= B(m2)\n\n\nQlatent= 334 kJ/kg(m2) =C(m2)\n\nWe have \n\nD(m1+m2)=A(m1)+B(m2)+C(m2)  you know that m1=.280kg  then you figure out m2? \nAnd \nA= 4190 (76.2 + 273)\nB=2100 (- 23.6+273)\nC=334,000\n\nActually Specific heat of ice = 2060 J/(kg*K) not 2100 J/(kg*K)\nAnd Specific heat of water  =4,183  J/(kg*K)
911974
Go with a line graph, that way you can also find out (or estimate) temperatures at times when you didnt measure it.
14789
A general form of the equation I beleive you are looking for comes from a basic understanding of thermodynamics.  I am assuming that your ct=(Tfinal-Tinitial)...\n\nSo a general form of the equation would be:\nmCp(Tf-Ti)=Q \n\n"Cp is the specific heat"\n\nWATCH UNITS!!!!!  \n(I am ignoring the conversions in this statement)\n55g*Cp(67degrees UNITS)=343 cal \n\nSolving this equation will give you Cp ... just make sure that you watch units.
722101
Surface Temperature = 5500ºC\n\nCore Temperature = about 15,000,000ºC
288780
From the following link, look for "Chapt22, Problem-24".\n\nThis problem is similar, except the speed is 10% instead of 11%. I'm sure you can use the same formulae though.\n\nhttp://www.jca.umbc.edu/~george/html/courses/phys112/2004fall/homework/04fall_phys112_hk8.pdf
149603
The amount of heat transfered is\n\n1) directlly proportional to the area of the sheet of given material. (A)\n\n2)directly proportional to the temperature difference on both sides of the sheet. (T2-T1)\n\n3)dirctly proportional to the time.(t)\n\n4)inversely proportional to the thickness of the sheet. (d)\n               \n---------------Q= A * (T2-T1) * t  /  d------------------------