This video made me so happy! The most helpful thing I have seen all year! I finally learned something this semester and feel confident about it! Thanks a lot man, keep making videos!
All is perfect. But in last example, you used a formula to calculate total heat transferred to hot reservoir as k=QH/W, where as you have introduced formula for coefficient of performance as K=QC/W in the first example. It creates confusion, which one is correct, if formula given in example 1 is correct then in example 3 the heat transferred to hot reservoir must be equal to 12500J. And heat taken out from cold reservoir equal to 10000J,
in the last example he used K=Qh/W because he was solving for a heat pump, whereas in the previous example when he used k=Qc/W, he was solving for a refrigerator.
how does the system behaves if we use the heat pump to recover heat from an exhaust fluid which is at a higher temperature than the heat sink. It is normally a spontaneous flow, and will not require energy input, but will instead output energy as work done. However, can we still use a heat pump in this case to increase the heat gained by the cold reservoir? For example to recover heat from some exhaust fluids which are at higher temperatures compared to the heat sink.
In a real life application of the heat pump, what temperature do we consider as that of the hot reservoir TH. In a heat pump for example, lets say we circulate water at 298K in the hot side(condenser of heat pump) which heats it up to 333K for home use. I guess Tc in this case would be the outside temperature, but which temperature in this case should we consider as TH?
Hi! Would you please consider doing a short video on the "Galilean Transformation"? Neil DeGrasse Tyson used it in a tweet today and I'm having trouble understanding it. Thanks!
I appreciate your videos but I find it very difficult to follow you when you write questions to be solve. The questions are never visible but other things are ok. Pls, can you write your questions out plainly for some one to see it clearly?. Thanks so much
Wouldn't the sign of the work be < 0? If work is done ON the system, then the sign of the work is positive because the system is gaining energy at the expense of the surroundings?
@@tusharsharma428 That means that the system loses energy when work is done on it: a system loses energy when the surroundings (an outside force), transfers energy into it. This is precisely the opposite of what happens.
Physics Video Lessons: www.video-tutor.net/physics.html
Bro I swear this guy made a video on everything. Such a life saver
Lol, you are right
You saved my ass in calculus and now with thermo, I love you.
I searched Heat Pump and Refrigerator on YT and first rec. video from this channel and thats>>>>>>>>>>>>>>>>>>>>>>>>>love you man
This video made me so happy! The most helpful thing I have seen all year! I finally learned something this semester and feel confident about it! Thanks a lot man, keep making videos!
my love for you can hardly be put into words
I would put money on the fact that im your biggest fan in the world right now.
lots of love from LDN
This video...... Man..... It's awesome. Crisp and to the point. Problem solving approach. That's what's needed. Great video.
He explains everything so comprehensively
best refrigerator video I can find, thank u so much!!
All is perfect. But in last example, you used a formula to calculate total heat transferred to hot reservoir as k=QH/W, where as you have introduced formula for coefficient of performance as K=QC/W in the first example. It creates confusion, which one is correct, if formula given in example 1 is correct then in example 3 the heat transferred to hot reservoir must be equal to 12500J.
And heat taken out from cold reservoir equal to 10000J,
in the last example he used K=Qh/W because he was solving for a heat pump, whereas in the previous example when he used k=Qc/W, he was solving for a refrigerator.
@@khaledsissi3762 how are both different? They work the same right?
Just fabulous explanation sir.... 🙏
how does the system behaves if we use the heat pump to recover heat from an exhaust fluid which is at a higher temperature than the heat sink. It is normally a spontaneous flow, and will not require energy input, but will instead output energy as work done. However, can we still use a heat pump in this case to increase the heat gained by the cold reservoir? For example to recover heat from some exhaust fluids which are at higher temperatures compared to the heat sink.
thank you!! You rock
Efficiency is 1-T2/T1=W/Q
W=w/k
P=Fv
what are you talking about shat is p and w/k
thank you very much for this. I hope I can pass the quiz for tomorrow.
did u end up passing?
Yes haha, luckily.
somehow, the voice is so good lol
Shukriya love it
Thank you so much sir.
Thanks
you saved my ass, again!
In a real life application of the heat pump, what temperature do we consider as that of the hot reservoir TH. In a heat pump for example, lets say we circulate water at 298K in the hot side(condenser of heat pump) which heats it up to 333K for home use.
I guess Tc in this case would be the outside temperature, but which temperature in this case should we consider as TH?
thanks bro
5:20
i like your voice
Hi! Would you please consider doing a short video on the "Galilean Transformation"? Neil DeGrasse Tyson used it in a tweet today and I'm having trouble understanding it. Thanks!
why in the first problem in the coefficient of performance formula you didn't multiply by 100% ? for part a)
Thank you! very useful video
I appreciate your videos but I find it very difficult to follow you when you write questions to be solve. The questions are never visible but other things are ok. Pls, can you write your questions out plainly for some one to see it clearly?. Thanks so much
How does this work with respect to the conservation of energy? it seems to cheat
Cop = t1 / ( t1 - t2 ) ???????
Hi can you do variables? Pls it's my birthday tomorrow
he did not do variables :(
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At which temperature ice cracking the steel tumbler inside the fridge 😰😁🔄🔀🌀⚡✴
JG
Wouldn't the sign of the work be < 0? If work is done ON the system, then the sign of the work is positive because the system is gaining energy at the expense of the surroundings?
formulas for internal energy diffrent in physics and chem
@@muneibrauf9779 Would you kindly give examples?
That is the case in thermo-d of chemistry......
In physics work done on the system is negative
@@tusharsharma428 That means that the system loses energy when work is done on it: a system loses energy when the surroundings (an outside force), transfers energy into it. This is precisely the opposite of what happens.
@@JRockTT yes, in the physical world we want machines to do work for us. That's why we consider it +ve.That's how I remembered it.
You need to show us your face, you saved me
Ur so sweet