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#66874 - aklidic
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#67133 to #66874 - copycopy
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(02/01/2013) [-]
Ahhh hi whats up?
#67154 to #67133 - aklidic
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#67155 to #67154 - copycopy
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(02/01/2013) [-]
Bordom trying to find somthing to occupy my time.
#67159 to #67155 - aklidic
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#67163 to #67159 - copycopy
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(02/01/2013) [-]
Oh I use to play violin but It ended up getting soaked in flood water so I had to throw it out been to broke to buy a new one. And I do I draw and play video games but it just seems to give me a headache at the moment.
#67166 to #67163 - aklidic
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#67167 to #67166 - copycopy
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(02/01/2013) [-]
Ahh would love to but can't seem to find the time to get to the library and find one that I enjoy. I havn't found much when it comes to literature since Inheritance. (Part of the eragon series.)
#67135 to #67133 - GeorgeBush
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(02/01/2013) [-]
copy :3!
#67137 to #67135 - copycopy
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(02/01/2013) [-]
George! Hi!
#67090 to #66874 - GeorgeBush
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(02/01/2013) [-]
There is a house hold heater that operates at 4 V and at 35 Ω and is used to heat up 15g of copper wire. The specific heat capacity of copper is 24.440 J/mol/K. How much time is required to increase the temperature from 25˚C to 69˚C?
#67098 to #67090 - aklidic
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#67118 to #67098 - aklidic
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#67104 to #67098 - GeorgeBush
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(02/01/2013) [-]
Unfortunately not, the answer comes out to be approximately 9.8 hours.

I'm really bored
#67117 to #67104 - aklidic
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#67132 to #67117 - GeorgeBush
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(02/01/2013) [-]
I'll try

Most of what that equation is lies in pluggin in information after you have the prompt. To find the power (P), you have to divide your volts (squared) by the resistance, P =V^2/R. So plug the numbers we know in:
P = (4)^2/(35)
P = (16)/(35), coming finally to P = .457 J/s

Now we need to find the equation by circuitry that will calculate power, which is q=mCsΔT*, with a note that "s" in that equation is meant to be subscript.

Now we plug in the numbers we know into the equation.

q=(15)(24.440)(69-25) where 15(g) is the total weight of the copper wire, 24.440(J/mol/K) is the cumulative heat capacity for said wire, and 69-25 is the difference in temperature. So,

q= (15)(24.440)(69-25) = 16130.4 J

So now we're left with 2 numbers, one tells us how many joules we must add to heat the metal, and the other tells us per second how many joules of energy the heater gives out. Now we just divide by the number of seconds. So,

(16130.4 J)/(.457 J/s) = 35296.3 seconds. 35296.3 seconds/60 gives us 588.27120 minutes. That, in turn, divided by 60 figures out to be 9.804530 hours.

#67148 to #67132 - aklidic
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#67152 to #67148 - GeorgeBush
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(02/01/2013) [-]
Even from GeorgeBush?
#67156 to #67152 - aklidic
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#66910 to #66874 - iwanttousenumbers
Reply +1
(01/31/2013) [-]
Are you single?
Are you single?
#66917 to #66910 - aklidic
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#66921 to #66917 - iwanttousenumbers
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(01/31/2013) [-]
Call me
#66876 to #66874 - aklidic
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#66908 to #66876 - reddeadtroll
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(01/31/2013) [-]
How's puberty?
#66916 to #66908 - aklidic
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#66885 to #66876 - shoryuken
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(01/31/2013) [-]
how's school?
#66891 to #66885 - aklidic
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#66911 to #66891 - shoryuken
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(01/31/2013) [-]
pretty good
#66893 to #66891 - anonmos
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(01/31/2013) [-]