A metal rod is placed between two temperature sources so that heat can flow between them. One temperature is $90.0^{\circ} \mathrm{C}$ and the other temperature is $25.0^{\circ} \mathrm{C}$. The length of the rod is $30.0 \mathrm{~cm}$ and the cross-sectional area is $10.0 \mathrm{~cm}^{2}$. If 15.0 watts of heat flows between the hot temperature and the cold temperature, then what is the thermal conductivity for the metal?

Question

Quizplus · Accepted Answer

The thermal conductivity ($k$) can be found using the formula $Q = \frac{kA(T_{hot} - T_{cold})}{d}$, where $Q$ is the heat flow rate (15.0 watts), $A$ is the cross-sectional area (converted to $m^2$, so $10.0 cm^2 = 0.0010 m^2$), $T_{hot} - T_{cold}$ is the temperature difference ($90.0^{\circ}C - 25.0^{\circ}C = 65.0^{\circ}C$), and $d$ is the length of the rod (converted to meters, so $30.0 cm = 0.30 m$). Solving for $k$ gives $k = \frac{Qd}{A(T_{hot} - T_{cold})} = \frac{15.0 \times 0.30}{0.0010 \times 65.0} = 69.2 \mathrm{~W} /\left(\mathrm{m} \cdot{ }^{\circ} \mathrm{C}\right)$.

A Metal Rod Is Placed Between Two Temperature Sources So 90.0∘C90.0^{\circ} \mathrm{C}90.0∘C

A Metal Rod Is Placed Between Two Temperature Sources So $90.0^{\circ} \mathrm{C}$