**Problem (IIT JEE 2004):**
Two identical conducting rods are first connected independently to two vessels, one containing water at $100\;\mathrm{{}^{o}C}$ and the other containing ice at $0\;\mathrm{{}^{o}C}$. In the second case, the rods are joined end to end and connected to the same vessels. Let $q_1$ and $q_2$ grams per second be the rate of melting of ice in the two cases, respectively. The ratio $q_1/q_2$ is,

- $1/2$
- $2$
- $4$
- $1/4$

**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}
{\Delta Q}/{\Delta t}=\kappa A {\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}
{\Delta Q_1}/{\Delta t}=2({\Delta Q}/{\Delta t})=\kappa A({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}
{\Delta Q_2}/{\Delta t}=\kappa A({100}/{2l}).
\end{align}
The heat transferred to the ice is used to melt it. The rate of melting is,
\begin{align}
q={\Delta m}/{\Delta t}=(1/L)\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., $q_1/q_2=\frac{\Delta Q_1}{\Delta t}/\frac{\Delta Q_2}{\Delta t}=4$.

**Problem (IIT JEE 1995):**
Three rods of identical cross-sectional area and made from the same metal form the sides of an isosceles triangle *ABC*, right angled at *B*. The points *A* and *B* are maintained at temperatures $T$ and $\sqrt{2}T$, respectively. In the steady state, the temperature of the point *C* is $T_c$. Assuming that only heat conduction takes place, $T_c /T$ is,

- $\frac{1}{2\left(\sqrt{2}-1\right)}$
- $\frac{3}{\sqrt{2}+1}$
- $\frac{1}{\sqrt{3}\left(\sqrt{2}-1\right)}$
- $\frac{1}{\sqrt{2}+1}$

**Solution:**
Let $T_A=T$, $T_B=\sqrt{2}T$, and the steady state temperature of *C* is $T_C$. Since $T_B > T_A$, heat flows from *B* to *A*. If $T_C*A* as well as from *B* leading to a non-steady state. Similarly, if $T_C>\sqrt{2}T$ then heat flows from *C* to *A* as well as to *B* leading to a non-steady state. Thus, in steady state $T*C* is equal to heat flow from *C* to *A* (see figure). The rates of heat flow from *B* to *C* and from *C* to *A* are given by,
\begin{align}
\label{poa:eqn:1}
&{\mathrm{d}Q_\text{BC}}/{\mathrm{d}t}={\kappa A(T_B-T_C)}/{x}={\kappa A(\sqrt{2}T-T_C)}/{x},\\
\label{poa:eqn:2}
&{\mathrm{d}Q_\text{CA}}/{\mathrm{d}t}={\kappa A(T_C-T_A)}/{\sqrt{2}x}={\kappa A(T_C-T)}/{\sqrt{2}x}.
\end{align}
In steady state, ${\mathrm{d}Q_\text{BC}}/{\mathrm{d}t}={\mathrm{d}Q_\text{CA}}/{\mathrm{d}t}$. Use above equations to get,
\begin{align}
T_C={(T_A+\sqrt{2}T_B)}/{(1+\sqrt{2})}={3T}/{(1+\sqrt{2})}.\nonumber
\end{align}

**Problem (IIT JEE 1991):**
A point source of heat of power $P$ is placed at the centre of a spherical shell of mean radius $R$. The material of the shell has thermal conductivity $K$. If the temperature difference between the outer and inner surface of the shell is not to exceed $T$, the thickness of the shell should not be less than ______.

**Solution**
The heat produced per unit time by the source is $P$. This heat is distributed over inner surface of the spherical shell having area $A=4\pi R^2$. Let $x$ be thickness of the shell. The rate of heat transfer due to conduction is,
\begin{align}
{\mathrm{d}Q}/{\mathrm{d}t}={K A T}/{x},
\end{align}
]where $T$ is temperature difference between inner and outer surfaces of the shell. For the temperature $T$ to be constant, the rate of heat incident on the inner surface of the shell should be equal to the rate of heat transfer through conduction i.e., $P={\mathrm{d}Q}/{\mathrm{d}t}$. Use above equation to get,
\begin{align}
x={KAT}/{P}={4\pi R^2 KT}/{P}.\nonumber
\end{align}