Question

Two identical conducting rods are first connected independently to two vessels, one containing water at ${100^o}C$and the other containing ice at ${0^o}C$. In the second case, the rods are joined end to end and connected to the same vessels. Let ${Q_1}$ ​ and ${Q_{2}}\;$ be the rate of melting of ice in the two cases respectively. Then the ratio $\dfrac{{{Q_1}\;}}{{{Q_{2}}\;}}$ is:
A. $\dfrac{{1\;}}{{2\;}}$
B. $\dfrac{{2\;}}{{1\;}}$
C. $\dfrac{1}{{4\;}}$
D. $\dfrac{4}{{1\;}}$

Answer 1

Heat will always flow from the hot body to a cold body. If we have one body with a higher temperature and the other body is colder than the other. Therefore heat from the hotter body transfers to the colder body. Heat energy flows to the air until the temperature of the hotter body and the surrounding air becomes equal. Therefore the heat flow is directly proportional to the temperature difference between the two.

Formula used:
$\dfrac{{dQ}}{{dt}} = KA\dfrac{{\Delta t}}{L}$
Here, $\dfrac{{dQ}}{{dt}}$ is the rate of flow of heat
$K$ is the thermal conductivity of the material.
$A$ is the area of cross-section.

Complete step by step solution:
Let us consider the rate of heat conduction through a material is having conductivity $K$, cross-section area $A$, length $\Delta x$ and temperature difference between two ends $\Delta x$ is given by,
Therefore we can write this as,
$\dfrac{{\Delta Q}}{{\Delta t}} = KA\dfrac{{\Delta T}}{{\Delta x}}.$
In case one the two rods are connected parallel. Also, it is given that the water’s temperature is ${100^o}C$ and the temperature of ice is ${0^o}C$. Therefore the temperature difference between these two is, $\Delta T$=${100^o}C$
Let us say that the two rods are separated by a distance, $\Delta x$=$l$
Therefore the rate of heat transfer through one rod is given by,$\dfrac{{\Delta {Q_1}}}{{\Delta t}} = KA\dfrac{{100}}{l}.$
Therefore through both the rods are, 2 \times KA\dfrac{{100}}{l}$$$$ = KA\dfrac{{200}}{l}
In case two the two rods are connected in series. Therefore the length between the two rods are $2l$
Now the rate of heat transfer to the ice is given by, $\dfrac{{\Delta {Q_2}}}{{\Delta t}} = KA\dfrac{{100}}{{2l}}$
The heat which is transferred to the ice is used to melt it. The rate of melting is given by how much energy is required to raise a mass of water by several degrees.
q = \dfrac{{\Delta m}}{{\Delta t}}$$$$ = \dfrac{1}{L}\dfrac{{\Delta Q}}{{\Delta t}}
Here$L$is the latent heat of fusion.
$\dfrac{{{q_1}}}{{{q_2}}} = \dfrac{{\dfrac{{\Delta {Q_1}}}{{\Delta t}}}}{{\dfrac{{\Delta {Q_2}}}{{\Delta t}}}} = 4.$
Therefore the correct option is D.

Note:
Latent heat is the amount of energy absorbed or released by a substance during a change in its physical state that occurs without changing its temperature. This latent heat associated with freezing a liquid or melting a solid is called heat of fusion. The latent heat is expressed as the amount of heat per mole or unit of mass of substance changing of state.