Visit ilectureonline.com for more math and science lectures! In this video I will show you how to calculate the final temperature of the water in a bucket when steam and ice are added.
hi my name is Zahra and I am 15 yrs old , and I am Saudi I just want to say that thank you very much for all your videos, here In Saudi we have a SAUDI team for the international physics Olympiad(IPHO) and I am part of it and all of my friend watch you all the time and find you very helpful I am very grateful,thank you
I worked half a year in calorimetry research studying the explosively exothermic polymerization of acrylic acid to make plexiglas at Rohm & Haas Research Division in Pennsylvania in the 1980s!
The first liquid temperature is 20°C Mass=m1 Heat capacity of the liquid is 4000J/kg°C The second liquid temperature is 80°C Mass=m2 Heat capacity of the liquid is 1000J/kg°C When those liquid are mixed together,Final temperature of the mixture becomes 30°C How to find m1/m2,Sir?
Hello Mr. Biezen, Thank you very much for the awesome videos, they are life-saving for students like me :) I have a similar problem and I would really appreciate it if you could help me understand it. In my problem I think the amount of steam is more than enough to melt all the ice, but I am having a hard time finding what the final equilibrium condition is (in terms of temperature and state of water). of course the final temp. will be 100 degrees C, but would the final state be a mixture of boiling water and steam at 100 degrees C? and if so, how much steam is present at equilibrium? the problem: 20 g of steam at 100 degrees C are added to 40 g of ice in a 30 g lead vessel, both of which are at -10 degrees C. determine the final equilibrium condition. note: specific heat of lead is 0.0306 cal/g.C.
It turns out that 68.5 % of the steam will turn into water at 100 degrees and 31.5 % of the steam remains as steam. The ice will melt and the melted water will also reach 100 degrees as will the lead.
Thank you very much Mr. Biezen. Now I know that I solved the problem correctly because I got a very approximate answer to yours. I am much more confident to get a good grade on my exam next week . Really can't thank you enough, you are the best :)
I'm a bit confused with the "+mL_v " ? I thought it would be - mL_v because I thought steam is losing thermal energy through condensation. It's going from a gas to a liquid so It should be -mL_v , its losing heat during the phase change??? Am I wrong? Although I do understand that there would be 2 terms because steam starts at 100C
It all depends on which equation you use to solve the problem. In this example we start with the equation heat gained = heat lost. Therefore all the terms on both sides of the equal sign must have a positive value.
How do I know when to use specific heat at kcal/ kg C with the L_f at kcal/ kg.... and when to use specific heat in J/kg with L_f's KJ/ kg? Does it matter as long as C's and L_f's units match?
Thank you for your time. If this isn't too much; would it be wrong to make the terms on the left side of the equation negative? I understand from your comments that it depends on the equation you use but can you clarify for this problem. For some questions using Qgained=Qlost becomes confusing when you have to factor out the Tf. Some terms are postive while others negative and it ends up giving a temperature far less than the predicted one.
There are indeed several ways in which you can approach this problem. The method shown here requires every term on both sides to be positive. From my experience, if students in my classes use the methods shown here, they often get the problem correct on the test. The students using the method shown in the book more often get the problem wrong on the test, thus I keep teaching this method. Although both methods are perfectly valid.
okay, I think I'll use your way as you're the only reason I haven't utterly failed mechanics. If you say that way is better then it must be. Thank you for the fast reply
That is a good question. I developed a new method to do that in the classroom, but haven't made any videos on that yet. (Maybe we can make them this weekend). Calculate each change separately on two sides: 1) How much energy it takes to heat the ice to 0 degrees. 2) how much energy it takes to melt the ice 3) how much heat it takes to heat that cold water to boiling on the other side 1) how much energy to remove to bring the temperature of the steam to 100 degrees 2) How much energy to remove to condense the steam. 3) How much energy to remove to drop the temperature of the boiling water to 0 degrees. When you look at those results side by side you can usually determine what the end state will be.
No, if you use this technique to solve calorimeter problems, every term on both sides of the equation must be positive. Thus you must subtract the lower temperature from the higher temperature. (Q gained = Q lost)
We have a playlist on that: PHYSICS 30 ENTROPY Physics - Thermodynamics: (1 of 5) Entropy - Basic Definition ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-MGA2uGck3iQ.html
Note that equation very specifically equates the heat gained by the cold objects to the heat lost by the hot objects. In this case all terms must be positive. Text books typically use a different method and do use positive and negative terms. You must use either the one or the other method, but you cannot mix them.
That is a good question. There are different methods to solve these types of problems. The method I use and show here in the video sets the heat (Q) gained by the cold objects equal to the heat (Q) lost by the hot objects. For the answer to be correct with this method all of the terms MUST be POSITIVE. In order to accomplish that you have to write T initial - T final if the T initial is greater than the T final.
ah yes, I see, were talking calories. very good. Thank you. I'm slightly lost in using Boltzman constant in thermodynamics, can you point me in the right direction? I understand it in entropy but it's relationship to R and in the idea gas law eludes me. Thank you for your time.
Just like any other equation in physics and nature, you need a constant to turn a relationship into an equation. In order to formulate the ideal gas equation you need a constant of proportionality. In that case it the the gas constant. See playlist: PHYSICS 26 THE IDEAL GAS EQUATION
You have to make an assumption of the final state. Let's assume that it will be water at a temperature between 0 and 100 C. Then you get the equation mc x delta(T) (steam) + mLv (steam) + mc x delta (T) warm water = mc x delta (T) ice + mLf (ice) + mc x delta (T) cold water
Sophia, When steam converts to water (condenses) there is no change in temperature. Therefore you only need the mass and the latent heat of vaporization
