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Mathematics 16 Online
OpenStudy (anonymous):

What is the relationship between power and temperature difference? How can you estimate power in a power vs change in temperature graph?

OpenStudy (anonymous):

Power is actually Energy/Time energy=mc Δt Power*time=mc Δt as the temperature change increases that means power also increases in other words energy per unit time increases.

OpenStudy (anonymous):

Power is directly proportional to (T)^4

OpenStudy (anonymous):

ok looking into it to verify

OpenStudy (anonymous):

well in this situation i have a 4.7ohm resistor, current is 1mA...not sure how you get what your saying...

OpenStudy (anonymous):

sory didn't reveal everything..kinda tired...hehe

OpenStudy (anonymous):

the experiment is a thermoelectric cooler connected to a power source and resistance

OpenStudy (anonymous):

first find the voltage v=ir

OpenStudy (anonymous):

the tec has 2 sides which have temperatures measured...but the voltage is collected from the tec so that varies in my graph of data collections

OpenStudy (anonymous):

power=IV

OpenStudy (anonymous):

multiply current * voltage and you will get the power of the source.

OpenStudy (anonymous):

yep thats right so at 65 degrees what would be the power if i am not given the voltage or the power...the voltage is measured through the tec

OpenStudy (anonymous):

i know that i is constant and the resistance is constant but it varies with the temperature otherwise the answer would always be the same for every temperature

OpenStudy (anonymous):

would i just find the gradient of the graph of previous collections, then find use y-y1=m(x-x1)? is that viable?

OpenStudy (anonymous):

can you upload the full question? Many things are jumbled up.

OpenStudy (anonymous):

ok i have to type it up... TEC has 2 sides. each side has a different temp which then makes \[DeltaT\] . R= 4.7 I=1mA There is a graph that is power vs \[DeltaT\] power and \[DeltaT\] data is collected every minute for 10 minutes. Estimate the power delivered to a 4.7ohm load by your TEC if the \[DeltaT\] is 65degrees

OpenStudy (anonymous):

ahhh there are spaces..

OpenStudy (anonymous):

the data in the graph really can be anything...so i won't post it as it is in an excel document

OpenStudy (anonymous):

okay so first you will be taking temperature every minute. Power from the tec will vary as the Δt will vary. The cooler will cool down the thing you are cooling right

OpenStudy (anonymous):

i guess what you need to do is not down the initial temperature and then find the final temperature where Δt = 65

OpenStudy (anonymous):

one side cools and the other side heats up

OpenStudy (anonymous):

it can be at any minute between 1-10 minutes.

OpenStudy (anonymous):

law of conservation of energy.

OpenStudy (anonymous):

if one side will lose about x j of energy the other side will gain x joul of energy.

OpenStudy (anonymous):

ahh change in t is actually to do with the hot and cold side

OpenStudy (anonymous):

ahh k

OpenStudy (anonymous):

and then we can find the change in temp by mcΔt=Q

OpenStudy (anonymous):

we weren't given mass or heat capacity

OpenStudy (anonymous):

in my notes i can see. change in E=Q-W

OpenStudy (anonymous):

and P=Q/t

OpenStudy (anonymous):

Pconduction =Q/t i mean

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