Two identical conducting rods are first connected independently: Heat Conduction
By Problem: Two identical conducting rods are first connected independently to two vessels, one containing water at 100°C and the other containing ice at 0°C. In the second case, the rods are joined end to end and connected to the same vessels. Let q1 and q2 grams per second be the rate of melting of ice in the two cases, respectively. The ratio q1/q2 is? (IIT JEE 2004)
- 1/2
- 2
- 4
- 1/4
Answer: The answer is (C) i.e., the ratio q1/q2 is 4.
Solution: The rate of heat conduction through a material having conductivity $\kappa$, cross-section area $A$, length $\Delta x$, and temperature difference between two ends $\Delta T$ is given by, \begin{align} \frac{\Delta Q}{\Delta t}=\kappa A \frac{\Delta T}{\Delta x}.\nonumber \end{align}
In case (1), two rods are connected in parallel.
The rate of heat transfer through each rod is, ${\Delta Q}/{\Delta t}=\kappa A ({100}/{l})$. Thus, the rate of heat transfer to the ice is, \begin{align} \label{vpa:eqn:1} \frac{\Delta Q_1}{\Delta t}=2 \frac{\Delta Q}{\Delta t}=\kappa A \frac{200}{l}. \end{align}
In case (2), two identical rods are connected in series making their effective length $2l$.
The rate of heat transfer to the ice is given by, \begin{align} \label{vpa:eqn:2} \frac{\Delta Q_2}{\Delta t}=\kappa A\frac{100}{2l}. \end{align} The heat transferred to the ice is used to melt it. The rate of melting is, \begin{align} q=\frac{\Delta m}{\Delta t}=\frac{1}{L}\frac{\Delta Q}{\Delta t}, \nonumber \end{align} where $L$ is latent heat of fusion. Use above equations to get the ratio of rate of melting in two cases i.e., \begin{align} \frac{q_1}{q_2}=\frac{\Delta Q_1}{\Delta t}\Big/\frac{\Delta Q_2}{\Delta t}=4. \end{align}
Related Question
Two metal cubes A and B of the same size are arranged as shown in the figure. The extreme ends of the combination are maintained at the indicated temperatures. The arrangement is thermally insulated. The coefficients of thermal conductivity of A and B are 300 W/(m-°C) and 200 W/(m-°C), respectively. After steady state is reached the temperature T of the interface will be? (IIT JEE 1996)
Answer: 60°C
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