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Basics of Conduction (in Hindi)
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This lesson discusses different modes of heat transfer and then explains the basics of conduction. It discusses mathematical expression for heat flow rate in conduction and illustrates with a few solved examples.

Shubham Kumar
I have done B.Tech, IIT Delhi (in Mechanical engineering). I have been teaching physics to JEE and NEET aspirants for three years.

Unacademy user
sir lcm and hcm lessions in RS Agarwal plz sir
before watching video can anybody tell me whether it has all laws of radiation or not
before watching video can anybody tell me whether it has all laws of radiation or not
Kya teaching style h Kabira tere desh m bhanti bhanti ke log
sir u helped me alot! jitna thnaks kru utna kam h... aapka teaching style best h. thank you once again...
sir from where 300 came in the ice box question...please reply sir
Simran Dahiya
7 months ago
Since it is given 5 minutes and 1 minute= 60 seconds. We multiply 5*60 and get 300.
  1. Heat Transter Three Modes: ConducHon: Hhen kere is a medium carryng heat and that medium is hastionary T, Convection When medium camging heat 4 in motion Hot Plate "> Radiation: when there ,w noinvolvement medium and heat is carrie^ by photon e g Radation of Sun reaching to easth

  2. Conducon 8. 2 2 Assumptions:- Heat leaving body at T = Heat reaching body at Ta "t There is no heat e behueen mediurm and Surssoundim "> The medium ; at teady state. This means heat is not being used to chmge temp of lhe medium

  3. emperature radient Let, lemp- Variation acron dx = dT 2 dx Temp grad- dT = Rate of change cf temp. with ength of medium If grad, is uniform.

  4. Rate of Heat Flo in Conducton 8 2 = Rate of heat Kaw = Heat 6Lowr p en unit time We may observe (A area of cro-secon) 2 conduckvity of matena Final equation Js in direchon decreatirg lemp)

  5. Example+ A rod of lengh m andro-sectional area 2 cm2 placed between a Annce &f constant "P 5d'c ond om ice box Containin ha of ice at 'c it taxes 5 mito melt alu th ice Hhen bind

  6. Solution ice 50 (Cons tomt 0 Magnitude of = provided to ce in o0 Heat 5 min = mx300 = 3K By Energy conservahor, = mLf 3K= 5003 80x4.2J

  7. Example Two voda of equal length. have rada us R 2R and Conduchwies 2K and K res pecti amanfed as hern in above tguretemp vely Ty ne

  8. Solution (g lo C Asuming no heat losses, nol C2nd yod) St -) 2 (60-T)= 41T-lo) 80 C 3 26.6 C