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Old Oct 6th 2013, 02:59 AM   #1
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Question regarding Heating effect of electric current and mechanical equiv. of heat

Well, I've encountered this problem as I was studying for my exams.
Basically, it's asking me to solve for the "mechanical equivalent of heat". I've scoured the internet for information regarding it, basically what I'm getting is that it's something like a conversion factor. So that makes me confused as to why it's making me look for something like that.

This set-up is quite familiar to me though, as I've encountered it in calorimetry and the concept where Q=mc(delta T) is involved.

I've also read some history on the mech equiv of heat, stating that it was some experimentally acquired constant, that's where i base my solutions from. I've also read that the values ranged from around 4.15 to 4.22.

So here's my take:
(from the heating effect off electric current)
H=V^2t/R (from's ohm's law)
H=16,500 J

(from that concept where Q=mc dT is involved)
Q1 = (153)(1)(35-10)
Q1 = 3825 cal

Q2 = (60)(0.1)(35-10)
Qtotal = 3975

(having prior knolwedge that the currently accepted value for J is around 4.19, I tried dividing the two answers)

J = 16,500J/3975Cal
J = 4.15 J/Cal
Am I spot on?
If I'm right, the question is like asking me to confirm the value of the constant am I right?

I'm still a bit confused about the "mech. equiv. of heat", is it just a conversion factor? Is there something more?
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Old Oct 6th 2013, 04:40 AM   #2
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I think you have it right. The notion of "mechanical equivalent of heat" dates from the early days of advances in thermodynamics in the 1850's, and came about with the realization that mechanical energy and heat are both two aspects of the same thing, and that it's really about conservation of energy and the specific heat of water. The notion that heat in calories is equivalent to mechanical energy in joules took some time to be accepted.
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