Latent Heat of Fusion and Vaporization, Specific Heat Capacity & Calorimetry - Physics

[@TheOrganicChemistryTutor]

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[@malikhind]

In the final problem you made a mistake. The change in T for Q(h2o) should be 27-21 not 27-20 and the answer should be 127 J/KG*C

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Suggestion plzz

[@yashjiwani1040]

127.5 to be precise

[@FirstClass_Chef]

I've tried to do this problem for more than 3 times getting that answer thinking there's something I'm not doing right

[@dailydoseofufc282]

@@yashjiwani1040 more like 127.4390244

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[@Lil-willski]

Same

[@Aem222]

Me too.

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Me three

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Q=ML -use whenever u have a phase change Q=mcAT -use whenever the temperature change

[@jadasarchives]

this video was so incredibly helpful, I understood everything after being lost for weeks in class. THANK U SO MUCH

[@christinarodriguez297]

A few tiny mistakes have already been mentioned below, but the video is otherwise EXTREMELY helpful!

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[@alishaislam9275]

Do u know how to do specific heat or heat of vaporization and heat of fusion I need help

[@omaralijimale5799]

@@alishaislam9275 Sorry I have seen this now

[@rosemacaskie]

When talking about the green house effect, people count vegetation as a dark material but vegetation is sculpturallly complicated, so that some bits of leaf are in the light and many others in shade so that they are often categorised as dark and so as collectors of heat not reflectors of visible light, when all sunlight falling on the reflective top of leaves get reflected back up skywards. If you are taking a photograph of leaves, the pools of light on the lit bit of the leaf annoyingly come out as a stronger shape than the actual leaf comes out as being, the shine coming off leaves often often comes out as light, as bits of white in the photograph it comes out as white. If you take a photo of the sunny side of a tree, its colour is much lighter than the other side but even this lighter side does not relfect the shinyness of leaves because on the sunny side of a plant you see the underside and so the non sunny side of the leaves that are above the camara and as leaves are often curved you also see the sides of leaves that are dark because curved away from the sun. I think all this should be mentioned when people categorise the effect of vegetation on climate change. Another aspect of the usefulness of vegetation as a coooler of the world, without mentioning how they take carbondioxide out of the air, is how very shinny hay and straw are, so this dead vegetation would, if left on the ground, both aislate what is a pretty good material as far as storing heat is concerned, earth, i have read that earth has the same ability to store heat as brick but also, hay and straw are very reflective grasses are full of glass, silica, which should put animals off eating grass, so hay and strawlike live leaves and probably more so, also reflect sun light back towards space. Have you thought how much land world over is left fallow for a year where wheat is grown? Half as much as is used to grow wheat, though this is not an agricultural habit used in modern Norht America? Also lots of land is ploughed in spring, just before the heat of summer, in preparation for autumn planting of wheat so leaving so much earth to accumulate heat, that it is absolultely mind boogling. once it did nto matter how much of a material which was a good accumulater of heat was left bare to be heated by the sun but now things have changed and it does matter.

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[@ANNIETOUCKEM]

Oh! Wow! this method of explaining is so amazing and very easy to understand. Teachers should explain like you. Thank you so much. 🙏🏼🙏🏼🙏🏼🙏🏼🙏🏼

[@xx-kid-yt2695]

30:58 the given initial temp for water is 21 degrees not 20 degrees. Thanks for the vid anyway, very helpful!

[@kabelokb902]

final answer is 127.62 J/kg C

[@manojar20]

26:48 for this question i got 24,094,500 J

[@juliuspahama3622]

Thank you so much for this vid. Now I undestand it well but there's a tiny mistake you made on the last problem. You wrote 27-20 instead of 27-21. Thank you soo much.

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Made me understand a whole section within few minutes...I appreciate your standard of explanation

[@bunganeradebe9984]

THANK YOU SIR FOR YOUR EFFORTS IN ASSISTING THE WORLD WITH PHYSICS I M FROM SOUTH AFRICA BY THE WAY, BUT AS I WAS DOING THE PROBLEM ON MY OWN I DISCOVERED THAT THE ANSWER IS 127.6 C . I THINK THE MISTAKE ON YOUR PART OCCURRED WHEN YOU WERE DETERMINING DELTA T , INSTEAD OF INITIAL TEMP 21 YOU HAD 20 DEGREES CELSIUS

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[@lilithasotondoshe8226]

Thank you very much for this video your explanations are always the best Sir. I think you made mistake at 29:50, final temperature of the water according to the given statement is 21 degrees Celsius not 20 therefore final answer of the problem is 127.6 not 148.9

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[@Haloly_Gamer]

the final answer for the specific heat of the metal is 127.6 J/kg*c , the mistake that you got at the end is the inital temp for H2O it is 21*c

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As I noticed too

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[@kareemteniola3634]

thank you so much for the explanation but there is a problem in the last question. you made a mistake somewhere at 27:02 if you check your calculations, it is heat gained by hot metal= heat gained by water but the problem is that when getting your temperature difference, you wrote (27-150) which is wrong. it is (150-27) because heat is being lost final temperature- temperature of the metal itself.

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WOW you explanation is perfect but in the last problem about the metal and water , in heat of water you compensated the change in temperature 7 but it should be 6

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Explain Latent heat of fusion, évaporation. Specific heat capacity, heat capacity and their differences as far Calorimetry is concerned

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how do you know to use the specific heat capacity of water or ice when dealing with something at 0 degrees?

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[@afolabikehinde6947]

There was a mistake in the final question. The initial temperature is 21 not 20. But good work though and thanks, the video was a lot helpful 👍💯

[@zainnunal7456]

In a physics class, the instructor has assigned a task to determine an experimental value for the heat of fusion of ice. The students dry and mass out 26.8-gram of ice and place it into a coffee cup with 100.0 g of water at 36.4°C. They place a lid on the coffee cup and insert a thermometer. After several minutes, the ice has completely melted and the water temperature has lowered to 18.1°C. What is their experimental value for the specific heat of fusion of ice? What is the answer?

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I guest there is wrong in multiplying 0.45 kg and 3.33x10E5 it must be 1,498,500 but anyways the process is correct and I learned a lot.

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[@ericamaedalayon1682]

Your channel always saves me from failing my exams and I'm thankful of it. But I think you are wrong at 10:34, because the specific heat capacity of water is 4.186 but instead you put 4186. You forgot to put decimal point between the 4 and 1 that's why the answer was huge.

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you had a mistake****

[@benitaiwobho6809]

The reason it's that way is because if you write it as 4.186 it'll be in grams, he multipied by 1000, to convert to kilograms, cause the questions carries that value...

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You are simply super amazing, even on the topic of the mechanisms of heat transfer you were great!!

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is there any video from a chemistry perspective covering the same topic?

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[@soos8649]

I think on the final problem you distributed your negative incorrectly for waters mcat value. q1+q2=0 so q1= (-q2) the negative is distributed over M, C, and Delta T. I get a specific heat of around 127.62 J/kg*degrees C. when I plug this specific heat value into q1+q2 I get around 0 (0.06=3767.4+(-3767.34)) and the values are rounded so that makes sense. If you plug in 148.9 to the metals specific heat you get the equation 0=3767.4+(-4395.528) which cant be. If i've made a mistake my bad

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