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100 amperes pass through a copper bar of $5$x$5$ mm cross-section. The resistivity of copper is $1.7 $x $10^{-8}$ ohm-metres.
Its volumetric heat capacity is $3.45$ joules per kelvin per cc.
Ignoring heat loss, what is the rate of increase of temperature of the copper in degrees C per second?

The question is pretty straightforward.
$R$$ =$$r\tfrac{l}{A}$
$Q$$ =$$I^{2}{R}{t}$

If change in heat per unit time$ = $$I^{2}{R}$,

temperature to cause this change in heat per sec is the answer .
But what is the equation that relates Temperature T and heat Q .
How can we come to answer to this problem ?

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closed as off-topic by John Rennie, JamalS, Pranav Hosangadi, Qmechanic Mar 3 '15 at 14:18

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It's given that volumetric heat capacity = c is 3.45 joules per kelvin per cc that is equal that the heat necessary to heat up 1 cc of cooper for 1 K is c, so $\Delta Q = CV \ \Delta T$ where $V = lA$ is a volume.

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  • $\begingroup$ $$ \frac{I^{2}r}{A^{2}{C}} $$ where $$r = resistivity $$ The answer should be 0.079 degrees/sec but I can't arrive there when I convert mm to m to calculate area A. Also 0.079 degrees/sec is in kelvin or celsius ? $\endgroup$ – coolstuff Dec 4 '14 at 19:37
  • $\begingroup$ The answer seems to be right. In this case there is no difference between Kelvin or Celsius because we use only change of the temperature but not the absolute value. $\endgroup$ – nvvm Dec 4 '14 at 19:52
  • $\begingroup$ @Jim but since area is in mm*mm , when we change to metres because C is J/K/cc, the answer is not 0.079, right ? $\endgroup$ – coolstuff Dec 4 '14 at 19:54
  • $\begingroup$ $$\frac{\mathrm{d} T}{\mathrm{d} t}=\frac{100*100*1.7*10^{-8}}{(5*10^{-3})^{2}*(5*10^{-3})^{2}*3.45}$$ does not give 0.079 degrees/sec $\endgroup$ – coolstuff Dec 4 '14 at 20:14
  • $\begingroup$ What does 5x5mm cross section mean ? Is it 5mmX5mm=25sq mm cross section area. $\endgroup$ – coolstuff Dec 4 '14 at 20:23
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Looking at the expression you wrote in the comment to @nvvm's answer, you are getting your units messed up. If you just convert everything to SI units, and write the units during your evaluation, you should see where your problem is.

You wrote

enter image description here

I would rewrite as

$$\begin{align}\\ \frac{dT}{dt} &= \frac{I^2R}{\ell\cdot A\cdot c_v}\\ &=\frac{I^2\left(\rho\frac{\ell}{A}\right)}{\ell\cdot A\cdot c_v}\\ &=\frac{100A \cdot 100A \cdot 1.7\cdot 10^{-8} \Omega\cdot m}{\left(25\cdot 10^{-6}m^2\right)^2\cdot 3.45 \cdot 10^6 J m^{-3}}\end{align}$$

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