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olga nikolaevna [1]
3 years ago
14

The resistivity of gold is 2.44 × 10-8 Ω · m at room temperature. A gold wire that is 1.8 mm in diameter and 11 cm long carries

a current of 170 mA. How much power is dissipated in the wire?
Physics
1 answer:
tester [92]3 years ago
6 0
Power in a wire where current is flowing can be calculated from the product of the square of the current and the resistance. Resistance is equal to the product of resistivity and length divided by the area of the wire. We do as follows:

Resistance  =  2.44 × 10-8 ( 0.11) / (π)(0.0009)^2 = 1.055x10^-3 <span>Ω

P = I^2R = .170^2 (</span>1.055x10^-3 ) = 3.048x10^-5 W
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A system dissipates 12 J of heat into the surroundings; meanwhile, 28 J of work is done on the system. What is the change of the
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Answer:

+16 J

Explanation:

We can solve the problem by using the 1st law of thermodynamics:

\Delta U = Q-W

where

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W is the work done (positive if done by the system, negative if done by the surroundings on the system)

In this case we have:

Q = -12 J is the heat dissipated by the system

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Substituting into the equation, we find the change in internal energy of the system:

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3 0
3 years ago
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pentagon [3]

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10500 J/kg/*C

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