22P3Q02

2a)

\displaystyle \begin{aligned}   (R&=\frac{\rho L}{A}) \\  \frac{R}{L}&=\frac{\rho }{A} \\  1.73&=\frac{\rho }{\pi {{(\frac{1.02\times {{10}^{-3}}}{2})}^{2}}} \\  \rho &=1.41\times {{10}^{-6}}\text{  }\!\!\Omega\!\!\text{  m} \end{aligned}

2bi)

Time after being switched on/s\displaystyle \Delta Uqw
0-59positivenegativepositive
80-100zeronegativepositive

COMMENT: Note that the wire is the system.

COMMENT: Since it is hotter than the environment, it is losing heat to the surrounding. Hence q is negative.

COMMENT: Remember that in electrical circuits, voltage (or potential difference) is defined as work done per unit charge? \displaystyle V=\frac{W}{q}. When applying the first law of thermodynamics to electrical systems, w refers to the energy that is being converted from electrical to non-electrical form, given by \displaystyle W=qV.

2bii)

\displaystyle W=Pt=VIt=(230)(12)(40)=110\text{ kJ}

\displaystyle \Delta U/Jq/Jw/J
0−110,000 kJ+110,000

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