anonymous
  • anonymous
I got my fingers burnt while working with a 12V, 7Amp. D.C. Battery. A copper wire that i was connecting to the negative, touched the positive terminal. What do i need to calculate the temperature of the wire if it took about 1 second to reach its maximum temperature (to become red hot). What is the formula?
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
Let's assume that: * the internal resistance of the battery and terminals is 0.1 ohms. * the heat exchange is zero (i.e. the heat did not transfer from the wire to environment) * the cross-section of the wire is 1 mm² and it's 10 cm long. First we calculate the resistance of the wire. Resistivity of copper is 16.78 nΩ·m and therefore the resistance of the wire is: \[R = \rho \frac{\ell}{A}=16.78\cdot 10^{-9}\frac{0.1}{1\cdot 10^{-6}}\approx 1.7m\Omega\] Then we calculate the current through the wire: \[I=\frac{U}{R}=\frac{12 V}{0.1 \Omega + 0.0017 \Omega}\approx118 A\] Then we calculate the power in the wire: \[ P=UI=RI^2=0.0017 \Omega\cdot (118 A)^2=24 W \] The specific heat capacity of copper is 384 J·kg−1·K−1 and the weight of the copper piece is 8.96×10^-5 kg therefore the temperature will rise \[ \Delta T = \frac{Q}{C}=\frac{Pt}{C}=\frac{24 W \cdot 1 s}{384 J·kg−1·K−18.96×10^-5 kg}=330 K \] 330 kelvins temperature increase will burn your fingers but it won't be red hot. Probably the resistance of the wire is a bit larger and therefore the power is larger and it will get hot sooner. And there are (at least) the following inaccuracies in my calculation: 1 The copper is not pure copper (resistivity might be wrong). 2 The resistance of the wire will increase when the temperature rises --> the power is not costant! 3 There is heat exchange between the wire and the environment. For example, the radiated heat will increase dramatically when the temperature increases. 4 The internal resistance of the battery + terminals is unknown and my 0.1 ohm was only an approximation.

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