@@GoodNewsForStrangers because they are teaching 20+ people who learn in different ways, they also need to go over multiple topics at a time. Imagine you have a 2 minute class and don’t pay attention for one minute (zone out) you’re done for!
Hello guys, just to help you guys out the equations is missing negative in the calculation. To end up with the Ans: 2.19 J/g °C i thought i would simplify for those that may need it. Don't forget the metal is losing energy and the water or liquid is gaining energy 'joules' So :- Qmetal/metal DT " -850J /(4.82) (-80.5°C) = 2.190665189. COOL WELL EXPLAINED VIDEO THOUGH. hope this helped.
At the beginning, it thought I would be really difficult, but you explained it well! I see that you omitted the cancelation of the negative signs of the metal q and the metal Delta temperature, that confused me at first but made me think a little harder (which was good!). Thank you again!
hello sir ! my self rishabh , i am from india. today fortunately i came to your channel. your style of teaching is fantastic it attracts me a lot . thanks for free education. you are making a great effort to make this world Literate wish you all the best. you cover the topics very earlier and effectivily it is very benificial for all
I would much prefer you use the equality of heat lost = heat gained. q(water) = -q(metal) Thus you can be consistent in the determination of delta T as final T - initial T. Your video is really nice for an introduction to the concept of calorimetry but may be confusing when instructors discuss exothermic changes which are given a negative q or delta H.
I know this video was a while ago but if you ever do remakes, putting everything in standard units would be awesome so it's consistent with college courses. Like kg, K, and all that. The visuals you put in are so helpful, thanks! ;]
For practical reasons, the change in water temperature is to be recorded in time. When the graphic peaks, that is the temperature change for the given equations. That is because the calorimeter change heat with the environment as well. Your thoughts?
At a village a mother heats 4 liters of water with a paraffin stove from a temperature of 18 °C. She later noticed that 20 g of the paraffin was used up. Calculate the final temperature of the water if the heat value of paraffin is 31 MJ/kg and the specific heat capacity of water is 4187 J/kg. °C.
My native language is Spanish but I study in Brazil so all my classes are in Portuguese and the only way that I found to understand this topic was theses videos in English hhahahahah, I love your videos
3:45 why was the difference in water temperature not subtracted from the temperature of the metal, and why the total temperature was subtracted from the temperature of metal?
i have long graduated and sometimes this very simple thing is not remembered (we cant remember everything right), so this video is an excellent video for a short revision.
Τhank you very much for sharing! I am watching you lectures to revise material which is needed in order to earn my master's degree. The only thing that I would like to say is that since the volume of the water has two significant figures, the end result should have two significant figures too [2.2 J/(gr*°C)] (3:38). Apropos, I couldn't find a metal that has this specific heat; could you reveal it to me?
I have another question is Does material has high heat capacity means to faster speed in change of temperature , in the condition of provide same heat to different material?
Hi Professor Dave THANK YOU! Your videos are very helpful with my SAT Chemistry prep. I have a question tho, how do you get delta T for the second equation of the metal?
Hi@@ProfessorDaveExplains, I also have a question. Why are we subtracting from the initial temp for the delta T (change in temp) and not from the final temp (i.e. 34.5 C - 115 C)?
Historical reasons. An early idea in defining the gram and kilogram was that milliliters of water would correspond to grams of water. This isn't strictly true with today's refined definitions, but it is approximately true enough that you can still get accurate calculations by assuming water's density is exactly 1 gram/mL.
Thanks that helped alot But did you assume that the metal started up at 25.5°C when the question doesn't mention it. I thought that when the starting temperature isn't mentioned we just take the standard which is 25. So my answer was slightly different, s≈1.958J/g°C
Please prof, I have a question on the enthalpy as, If the elementary step A to B has a reaction enthalpy of XKJ and the activation energy of the reverse step B to A is YKJ, then the activation energy of the process A to B would be what?
Hello prof. , im confused by the concept of this question which is contrast to my understanding, the solution explains that the q of the diamond is equal to the "-" q of the water (q[diamond]= - q[water]. Why did the water absorbed heat and has a negative value? 5. As a purity check for industrial diamonds, a 10.25-carat diamond (1 carat = 0.2000 g) is heated to 74.21 oC and immersed in 26.05 g of water in a calorimeter. The initial temperature of the water is 27.20 oC. If the specific heat of diamond is 0.519 J/g oC, what is the final temperature of the water?
For the comprehension question: In the question itself, you said that the unknown metal was placed in 35mL of water. However, when solving for the water, you wrote 35g. Was that a mistake or was it intentional?
These videos are very helpful. But just one suggestion, put them into an organized playlist of some sort so that people don't have to go around and search for the next topic
Hi, umm.. I had a question about the metal and water example... When you were calculating the specific heat of the metal, why did you choose the ΔT to be the difference of temperature between the temperature of the boiling water and temperature of the water on coffee cup?
In the problem 4.82g of an unknown metal is heated to 115.0c and then placed in 35L of water at 28.7c,which then heats uptp 34.5c.What is the specific heat of the metal? For the waterq=(4.186 j/g) (35g)(5.8c) how or where did you get the 5.8cand for the Metal 850j =s(4.82g)(80.5c) S=2.19 j/g cwhere or how to get 80.5c as these temperatures are not mentioned in the question and also when i multiply the specific heat and the temperature i dont get 2.19 j/g ci get 4.82*80.5=388.01 Please explain at your earliest convenience as iam in a dark dark place
Mathematically you can use grams, but the SI unit of mass is the kg. So specific heat capacity is often quoted in joules per kg per degree C. (Wouldn't it be nice if the SI unit was gram)
hey professor dave, thanks a lot for the explanation but I have a question though in the example how could the initial temperature of the metal be 100c and the water is 25c but the final temperature for both of them is 33.8c shouldn't it be at 62.5c so both of them can be at thermal equilibrium? are these just arbitrary numbers for the example or am I missing something??
the hot metal starts at the hot temperature, and it is placed into room temperature water. then the system equilibrates at an intermediate temperature, but these objects have totally different specific heats, the metal cools at a different rate than water heats, so the final temperature is not just the average of the temperatures, you have to do the math
If amount of water is given in milliliters, then you have to translate to mass through the density of water. For water, for what it's worth, 1 mL is approximately 1 gram. That is, unless you have volumetric heat capacity values available to you.
deltaT = final temp - initial temp for the metal, delta T= 34.5- 115.0 = -80.5 it is a negative number. why was a positive value put in for calculation?
actually I think I got the answer to my own question. q(absorbed) should equal to negative q(released). so in solving for the metal, it should be -850 J= s (4.82g)(-80.5 degree Celsius) s=2.19 (a positive value) is this correct?
It's my 2nd semester of first-year university chemistry... and of course, Professor Dave saves me once again.
Same here and agreed 🥹
@@yunyung lmao same as well
Then why am I doing this in 10th grade. Not fair
Then why am I doing this in 10th grade. Not fair
@@riruruuu bcaz what that person is doing is a intro chem course, they refresh the basics first then dive into the harder stuff as they progress
"He knows lot about science stuffs, Professor David explains! Turuttu ! "
Lol
It's dave*
@@DvolWOLF no it's chemistry Jesus
I love that too lol I'm always singing along 😂
How tf does that sound like "turuttu" to you ?
you are amazing
what the teacher do in 50 min
you did it in 5 min
I always wondered why it would take them 30 minutes to explain a concept which can be done in 2 minutes.
@@GoodNewsForStrangers because then they wouldn't have jobs lol
اا والله ^_*
indeed
@@GoodNewsForStrangers because they are teaching 20+ people who learn in different ways, they also need to go over multiple topics at a time. Imagine you have a 2 minute class and don’t pay attention for one minute (zone out) you’re done for!
Hello guys,
just to help you guys out the equations is missing negative in the calculation. To end up with the Ans: 2.19 J/g °C i thought i would simplify for those that may need it.
Don't forget the metal is losing energy and the water or liquid is gaining energy 'joules' So :- Qmetal/metal DT " -850J /(4.82) (-80.5°C) = 2.190665189.
COOL WELL EXPLAINED VIDEO THOUGH.
hope this helped.
this was the one thing i was confused about, thank you so much
At the beginning, it thought I would be really difficult, but you explained it well! I see that you omitted the cancelation of the negative signs of the metal q and the metal Delta temperature, that confused me at first but made me think a little harder (which was good!). Thank you again!
hello sir ! my self rishabh , i am from india. today fortunately i came to your channel. your style of teaching is fantastic it attracts me a lot . thanks for free education. you are making a great effort to make this world Literate wish you all the best. you cover the topics very earlier and effectivily it is very benificial for all
Same bro
I didn't do the comprehension I was to busy jamming to the sick beats
check out the song at the end of any clip in my organic chemistry practice problems series. that's the dopest beat of them all.
Wow you're right I'm adding it to my playlist rn ✌🏾
I would much prefer you use the equality of heat lost = heat gained. q(water) = -q(metal) Thus you can be consistent in the determination of delta T as final T - initial T. Your video is really nice for an introduction to the concept of calorimetry but may be confusing when instructors discuss exothermic changes which are given a negative q or delta H.
Thanks Professor Dave for doing something in four minuets my professor couldn't do in forty minuets. Explain the material.
Our course never differentiated heat capacity and specific heat so this video clears up so many questions I had 🎉
it took you 4 mins to explain an hour lecture from my professor. thank you
thank you so much jesus i have a test tomorrow and you came in clutch ❤❤❤❤❤❤❤❤
I know this video was a while ago but if you ever do remakes, putting everything in standard units would be awesome so it's consistent with college courses. Like kg, K, and all that.
The visuals you put in are so helpful, thanks! ;]
so u are a studend of thermodynamics?
Yes, it is better to use international units. This is more standardized and less likely to cause ambiguity.
This dude is so amazing even the teacher all over the philippines LINKED his vids for me and my class mates to whatch
Thankyou for the video on Heat Capacity, Specific Heat, and Calorimetry.-Alexis Kironde.
Such a wholesome comment written like an email
This is TH-cam you didint have to be that formal take a chill pill😂
How do I always get suggestions on what I'm currently learning in class!? 😮 Professor Dave is psychic!
What is the initial value of metal temperature?
For practical reasons, the change in water temperature is to be recorded in time. When the graphic peaks, that is the temperature change for the given equations. That is because the calorimeter change heat with the environment as well. Your thoughts?
This really helped my understanding of this topic thank you!
Awesome video hit the nail right on the head
Why is 66.5 degrees not a negative change in temperature at 3:05? The temperature of the metal decreases to reach the final temperature, right?
What would the research question for the coffee cup experiment be?
thank you jesus
😂
Thank you Professor Dave!!!!!
At a village a mother heats 4 liters of water with a paraffin stove from a temperature of 18 °C. She later noticed that 20 g of the paraffin was used up. Calculate the final temperature of the water if the heat value of paraffin is 31 MJ/kg and the specific heat capacity of water is 4187 J/kg. °C.
I am from Bangladesh. Really your lecture help me.
My native language is Spanish but I study in Brazil so all my classes are in Portuguese and the only way that I found to understand this topic was theses videos in English hhahahahah, I love your videos
Really? kkkkkkk I'm from Brazil
lovee the introoo
i already start singing it when i click on ur video
thank you brother, u help alot of fellow studs by ur lecs
GOD BLESS YOU!!
3:45 why was the difference in water temperature not subtracted from the temperature of the metal, and why the total temperature was subtracted from the temperature of metal?
I like how simply Professor Dave explains science stuff
I love you Professor Dave
Very intelligently explains. Prof Dave is the best
I love this man!
Thank you so much, your amazing! ❤️
Wow, I needed that visualization. 👀👌 This is good stuff.
I love your content so much but bro you your intro is the funniest thing ever it really made my day
i have long graduated and sometimes this very simple thing is not remembered (we cant remember everything right), so this video is an excellent video for a short revision.
Τhank you very much for sharing!
I am watching you lectures to revise material which is needed in order to earn my master's degree.
The only thing that I would like to say is that since the volume of the water has two significant figures, the end result should have two significant figures too [2.2 J/(gr*°C)] (3:38).
Apropos, I couldn't find a metal that has this specific heat; could you reveal it to me?
The best part is your video is concise and useful . thanks
2:31 Is the specificheat before or after the water absorbs the heat from the rock?
3:45 Why is the metal's temperature variation 80.5 °C? Shouldn't it be 5.8°C?
I'm confused too
Very insightful helped me to understand that coffee cup experiment
Amei a explicação ❤🎉 muito obrigada! O Brasil te assiste!!
I'm confused, how did you calculate the specific heat of an object?
Me too 😢😢😢
THANK YOU !
I have another question is
Does material has high heat capacity means to faster speed in change of temperature , in the condition of provide same heat to different material?
to think this the same dude who be cussin out flat earthers i love this guy frl
Hi Professor Dave
THANK YOU! Your videos are very helpful with my SAT Chemistry prep.
I have a question tho, how do you get delta T for the second equation of the metal?
the initial temperature of the metal is 115 so you have to use that!
@@ProfessorDaveExplains sir why we are taking the subtraction whether both are differently placed?
Hi@@ProfessorDaveExplains, I also have a question. Why are we subtracting from the initial temp for the delta T (change in temp) and not from the final temp (i.e. 34.5 C - 115 C)?
What's the difference between ''heat transfer'' and ''change in temperature''?
Really helped subscribe to this channel
do it
Dave, you are really good.Thank you very much.
Why does the mL of the water equal the grams of water? How do you come to that conclusion?
Water is 1.00 g/mL at room temperature.
Historical reasons. An early idea in defining the gram and kilogram was that milliliters of water would correspond to grams of water. This isn't strictly true with today's refined definitions, but it is approximately true enough that you can still get accurate calculations by assuming water's density is exactly 1 gram/mL.
You are very fantastic prof. Thank you for making this channel
😊😊👍👍👍
I am Indian Sir ...really your explanations are better than our teachers..continue making videos sir
Thanks that helped alot
But did you assume that the metal started up at 25.5°C when the question doesn't mention it. I thought that when the starting temperature isn't mentioned we just take the standard which is 25. So my answer was slightly different, s≈1.958J/g°C
thank you professor Dave, for eplaining me calorimetry in simple words.
Please prof, I have a question on the enthalpy as, If the elementary step A to B has a reaction enthalpy of XKJ and the activation energy of the reverse step B to A is YKJ, then the activation energy of the process A to B would be what?
I will do anything to make u my physics teacher....Thanks for these amazing videos !!!!!!! 🤩
He looks like Ranveer Kapoor
Hello prof. , im confused by the concept of this question which is contrast to my understanding, the solution explains that the q of the diamond is equal to the "-" q of the water (q[diamond]= - q[water]. Why did the water absorbed heat and has a negative value?
5. As a purity check for industrial diamonds, a 10.25-carat diamond (1 carat = 0.2000 g) is heated to 74.21 oC and immersed in 26.05 g of water in a calorimeter. The initial temperature of the water is 27.20 oC. If the specific heat of diamond is 0.519 J/g oC, what is the final temperature of the water?
That last q helped me understand that I understood. Tnku
thank u very much really its clear more about calorimeter
For the comprehension question:
In the question itself, you said that the unknown metal was placed in 35mL of water. However, when solving for the water, you wrote 35g. Was that a mistake or was it intentional?
water is 1.0 g/mL!
Your videos r very useful but can u plz talk about thermal decompostion
I need some help. Could someone provide me with the specific heat of caramel color? If possible, please send a document. Thanks.
How did u change the volume into mass
Hi, how do you get the change in temp of metal equal 66.2 ) at 2:49 min? thank you
Did you take 100 degree as initial temp of metal and final temp is after the metal temp equilibrate with liquid at 33.8 degree?
yep that's right
Each of your videos first illuminates my mood and then your reference illuminates my concentration power.😇
sir, will you define heat capacity with some other lab activities??
Specific Heat of Water is 4.186 J/gC for the comprehension part.
TheChosenOne thanks. i was going crazy as i was using 1 cal/g C and getting the wrong answer hahah
how does he get 4.186?
@@keylal3339 exactly i want to know as well
Awesome comprehension question!
It was a couple years I haven't learn about different of heat capacity or specific heat. Thanks for explaining us.
I am from Egypt , Really Thnx 💙
These videos are very helpful. But just one suggestion, put them into an organized playlist of some sort so that people don't have to go around and search for the next topic
buddy all of my tutorials are immaculately organized into playlists! did you even think to check before commenting?
Sir, how specific heat different from conductivity or with other mode of heat transfer....
Specific heat is not a means of heat transfer , it is the characteristic property of every substance and is different for all.
Hi, umm.. I had a question about the metal and water example... When you were calculating the specific heat of the metal, why did you choose the ΔT to be the difference of temperature between the temperature of the boiling water and temperature of the water on coffee cup?
So the metal starts out in the boiling water, and it ends up in the cup at equilibrium, so those are the two temperatures for the metal!
The best explainer❤
where did he get the 9.25g and 66.2 C in his computation for specific heat of metal?
2:12 how did you get the final temp?
reading on the thermometer
I’m literally watching this 2 hours before my AP chem exam. 😂
I'm watching it during an exam lol
You should probably change the definition of specific heat capacity at the start to 1kg not 1 gram.
very nice explained sir
very great video! prof dave
i got more confused but thanks
for the comprehension part don't you have to convert ml --> grams because "m" is for grams in he equation?
nvm
Professor, I really didn't understand that from where did we get the value (80.5°C) in the comprehension ???
Someone please help me...🙏🏻
it's just the change in temperature!
But how did you work that out
@@kevinakussah4736 final temperature minus initial temperature
Liked and subscribed. Thank you professor Dave, this is dope
why it is not -80.5 degree celcius as dell T is T final - T initial
∆T = T final - T initial
In the problem 4.82g of an unknown metal is heated to 115.0c and then placed in 35L of water at 28.7c,which then heats uptp 34.5c.What is the specific heat of the metal?
For the waterq=(4.186 j/g) (35g)(5.8c)
how or where did you get the 5.8cand for the Metal 850j =s(4.82g)(80.5c)
S=2.19 j/g cwhere or how to get 80.5c as these temperatures are not mentioned in the question and also when i multiply the specific heat and the temperature i dont get 2.19 j/g ci get 4.82*80.5=388.01
Please explain at your earliest convenience as iam in a dark dark place
I am preparing for IIT😊
For the "Checking Comprehension" question, it uses 35mL for water but you used 35g in the solution. Is that interchangeable here or a typo?
water is 1.0 g/mL so yep they are interchangeable! only for water though.
I have my examen in 1 hour and I so nervious, thanks for help me
shouldn't be the q in computing for the specific heat capacity of metal be negative? Heat was released by the metal isn't it?
Mathematically you can use grams, but the SI unit of mass is the kg. So specific heat capacity is often quoted in joules per kg per degree C. (Wouldn't it be nice if the SI unit was gram)
hey professor dave, thanks a lot for the explanation
but I have a question though
in the example how could the initial temperature of the metal be 100c and the water is 25c
but the final temperature for both of them is 33.8c
shouldn't it be at 62.5c so both of them can be at thermal equilibrium?
are these just arbitrary numbers for the example or am I missing something??
the hot metal starts at the hot temperature, and it is placed into room temperature water. then the system equilibrates at an intermediate temperature, but these objects have totally different specific heats, the metal cools at a different rate than water heats, so the final temperature is not just the average of the temperatures, you have to do the math
oh I didn't know that my teacher just taught it to me like that in prep school or something :""
I appreciate the very quick response too! thanks a lot
Sir specific heat capacity can we called as specific heat or both differ??
nope they are different! heat capacity applies to a whole object, specific heat applies to one gram of a particular material.
But specific Heat is the same thing as specific heat capacity because all your adding is a capacity to the name.
@@dennisc5607 ah i see, yes those are the same.
Professor Dave Explains 👍
Quick question. How are you supposed to do the equation of water is in milliliters?
If amount of water is given in milliliters, then you have to translate to mass through the density of water. For water, for what it's worth, 1 mL is approximately 1 gram.
That is, unless you have volumetric heat capacity values available to you.
deltaT = final temp - initial temp
for the metal, delta T= 34.5- 115.0 = -80.5 it is a negative number. why was a positive value put in for calculation?
actually I think I got the answer to my own question. q(absorbed) should equal to negative q(released). so in solving for the metal, it should be
-850 J= s (4.82g)(-80.5 degree Celsius)
s=2.19 (a positive value)
is this correct?
@@cyoon87 thx bro i was confused about it too
@@cyoon87 yeah probably, I was confused but that cleared it up. It’d still be positive
His face looks like ranveer Kapoor
No?
Slightly same like him
You are the desi indian
Too much mouth spitting
No man like annat ambani
How did you come up with 80.5 Celsius ?
lovely explaination
Respectfully forgot the proper sig digs
Good catch man
please respond would make my year brotha