The equation is valid during steady state or equilibrium conditions or instantaneously valid under non-equilibrium conditions. The way to look at this is to focus on the boundary between the two materials that are responsible for the conductive heat transfer. Looking at the boundary itself between the two materials and assuming the hotter bath is on the left side, the equilibrium or steady state condition means that the heat transfer coming from the left side across the boundary must match the amount of heat received on the right side of the boundary. This is to be expected - the head transfer across the boundary is steady. This will only be true under equilibrium conditions (or instantaneously otherwise) and the temperature of the conductive material will (at steady state) be constant. Prior to equilibrium and steady state, the conductive material's temperature will rise... Under equilibrium conditions, the heat bath on the left is at a steady temperature T1 and transfers heat via conduction across the connecting material to the heat bath on the right, which is itself at a steady temperature T3. The conductive material, under steady state reaches a temperature between the hot bath at T1 and the cold batch at T3. Under these conditions, all temperatures are steady and heat transfer is steady. T1>T2>T3. This model also assumes that the source of heat is 'infinite' eg that the hot bath is not cooling down over time as heat is transferred out of it or that this transfer is small relative to the size of the bath and hence can be ignored - if the hot bath cannot be treated as infinite, then, as heat is transferred out of it, the hot bath will cool down and all three materials - left bath, conducting connective material, and right bath, will all hit a fixed, final temperature.
It is similar to water flowing through a hose. If you pick any point inside the hose (call it a junctions), the amount of water flowing into that point must equal the amount of water leaving that point. Or think of an intersection. The amount of cars entering the intersection must equal the amount of cars leaving the junction
I'd first like to say thanks for all your series they are amazing. I understood how you solved it , and everything calculation wise . However i cannot understand get one thing straight. I can imagine two metals with metal rods of same diameter sticking out from each other. I would imagine that at the instant of them being connected the temperature at the contact would initially be split into 100 C and 0 C and the would slowly change its temperature to reach an equilibrium value as a function of time depending on the rate of Q. I don't understand which part is 95 C and what metal this is , at which instant of time Thanks in advance
Thanks for your great video, it is really very helpful for me ! do you have any example which is combine the conduction and convection together? I have a heat power like 5W on one left face of a rectangle block, I want to know the temperature distribution along the block after I put the heat power with the consideration of both convection and conduction, can you give me some suggestion on how to calculate it? Thanks!
isnt ∆T supposed to be Tf - Ti? if so, then 100°c is the initial temperature of the copper to the junction(that has the final temperature) and the ∆T in Copper must be = (T - 100)?
so, just to clarify, this setup is such that neither of the sides (100C and 0C) ever change temperatures, so each point on the pipe will find an "equilibrium" of a certain temperature they will forever stay at?
Hi why isn’t it T final minus T initial in the equations on both sides ? For one you did T final minus T initial but for the other side you did it the opposite way.
I don't understand why did you inverse the expression of delta T, I've been told that it's aqual T final-T initial which means (T-100) and (0-T) not what you wrote. So could please clarify that part for me Sir. And thanks for your efforts.
We are just looking for the magnitude of the temperature difference. Since we already know where the hot side is and where the cold side is, we don't have to worry about the sign of the delta T, just the magnitude.
The negative is used to relate the direction of flow to the positive or negative change in the temperature. Since we know that the heat flows from the hot reservoir to the cold reservoir, and we take the delta T as a magnitude, we don't need the negative sign.
It is like a water hose. For every drop of water added on one side, one drop will come out on the other side. And throughout the hose 1 drop will be moving along the hose. The same thing happens to current on a wire.
I have the same physics problem but this time the 2 temperatures of both the boiling water and the cold water are missing, the question didn't have the length for the L2 either, so there are 3 missing variables.I was told to find out the heat flow to the copper section which equal to the heat flow to the iron section, which finally equal to the heat flowing to the ice water. Please help!
A long rod, insulated to prevent heat loss along its sides, is in perfect thermal contact with boiling water (at atmospheric pressure) at one end and with an ice-water mixture at the other. The rod consists of a 1.00m section of copper (with one end in the boiling water) joined end-to-end to a length L2 of steel (with one end in the ice water). Both sections of the rod have cross-sectional areas of 4.00 cm2. The temperature of the copper-steel junction is 65.0∘C after a steady state has been reached. Assume that the thermal conductivities of copper and steel are given by kcopper=385Wm⋅K and ksteel=50.2Wm⋅K.
The question asked for the amount of heat per second H (=QΔt) flows from the boiling water to the ice-water mixture? and how long id L2 (steel section)?
this doesnt make sense...how does the material NOT affect the rate of heat flow? in the previous video you said that depending on the material and (k) rate of heat is faster/slower respectively
I don't quite understand your question since heat flow IS affected by the material. Both coefficients of heat conductivity of the 2 materials are defined and used in the problem.
sorry , I meant how can dq/dt be equal for both materials , since for example Copper has better heat conductivity than Iron (I think) thus dq/dt would be bigger while the heat was moving on the copper part , no?
Ah now I understand the question. It is like a pipe carrying water. The amount of water flowing through a pipe remains constant even when the diameter of the pipe changes (the velocity changes to accomplish that). The same can be said about heat flow. (You cannot have more heat flowing in than flowing out of a section of conductor).
right but how do we explain this algebraically? like if we divide both dq/dt for the different k of the conductors we dont get 1, since the k constants are different and they wont cancel out. Sorry for being so extensive about this, also thanks for the replies
solve for the equation of each piece separately in terms of dQ/dt. Then set the 2 dQ/dt's equal to each other (because you know they must be the same).
how can I find the temperature at some distance x from one of the edges as a function of time, given the initial temperature of the rod, and assuming that the rod is uniform
Greetings, I don't know if this will get a response, but I have a question. If we had multiple materials (of course with same cross sectional area) of different lengths, would it be possible to equate the following?: (sum of all k's) (delta T) / Length = (sum of other k's) (delta T) / Length ? I would appreciate any response
I would start with dQ/dt for 1 = dQ/dt for 2 = dQ/dt for 3 etc. (The amount heat flowing through each section is the same) Provided they are connected in series (not parallel)
In this case it is not necessary because the units of length cancel out. If you are not sure, it is always better to convert to make sure you don't make any errors.
k-copper>>k(fer) = ( 397>>80)= .It means conductivity coeff of copper MUCH BIGGER than Fe/iron.-----Cu = very good for heat transfer ( poor for insulation)\
If the hot and cold reservoirs are kept at the same temperature, then the amount of heat flowing through the bar must be constant. You cannot have more heat flow through one part of the bar than the other part of the bar. (where would the extra heat go to?)
The same amount of heat that flows through the left side of the path, must be the same as the amount of heat that flows through the right side of the path. (It is similar to the amount of water flowing through a pipe, you cannot have more water flowing through one section and less through another section).
+Binrio “If You See Kay” Biñan A represents the cross sectional area of the Cu and Fe bars and they are the same for both. (So we divide both sides of the equation by A)
