Question

There are two hollow spheres made of different materials, one with double the radius and one-fourth wall thickness of the other are filled with ice. If the time taken for melting the ice completely in a larger sphere is 25 minutes and that for a smaller sphere is 16 minutes, find the ratio of the thermal conductivity of the larger sphere to the smaller sphere.
A. 4:5
B. 5:4
C. 25:1
D. 1:25

Answer 1

Hint:In order to solve this problem we need to understand the rate of heat transfer. The rate of flow of heat is the amount of heat that is transferred per unit of time. Here, using the formula for heat flow we are going to find the solution.

Formula Used:
To find the heat flow the formula is,
\[\dfrac{Q}{t} = KA\dfrac{{\Delta T}}{L}\]
Where,
A is a cross-sectional area.
\[\Delta T\] is the temperature difference between two ends of the metal.
L is the length of the metal plate.
K is the thermal conductivity.

Complete step by step solution:
The quantity of heat is nothing but how much the heat flow per unit of time in a given material and is given by,
\[Q = KA\dfrac{{\left( {\Delta T} \right)t}}{L}\]
\[\Rightarrow K = \dfrac{{QL}}{{A\left( {\Delta T} \right)t}}\]
Since Q and \[\Delta T\] are the same for both spheres.
Then,
\[K \propto \dfrac{L}{{At}}\]
\[\Rightarrow K \propto \dfrac{L}{{\pi {r^2}t}}\]
\[\Rightarrow K \propto \dfrac{L}{{{r^2}t}}\]

For larger spheres,
\[{K_{larger}} = \dfrac{{{L_l}}}{{{r_l}^2{t_l}}}\]
For smaller spheres,
\[{K_{smaller}} = \dfrac{{{L_S}}}{{{r_S}^2{t_S}}}\]
Here, the ratio of the thermal conductivity of the larger sphere to the smaller sphere is
\[\dfrac{{{K_{larger}}}}{{{K_{smaller}}}} = \dfrac{{\dfrac{{{L_l}}}{{{r_l}^2{t_l}}}}}{{\dfrac{{{L_S}}}{{{r_S}^2{t_S}}}}} \\ \]
\[\Rightarrow \dfrac{{{K_{larger}}}}{{{K_{smaller}}}} = \dfrac{{{L_l}}}{{{r_l}^2{t_l}}} \times \dfrac{{{r_S}^2{t_S}}}{{{L_S}}}\]
\[\dfrac{{{K_{larger}}}}{{{K_{smaller}}}} = \dfrac{{{L_l}}}{{{L_S}}} \times \dfrac{{{r_S}^2}}{{{r_l}^2}} \times \dfrac{{{t_S}}}{{{t_l}}}\]
Here, \[{r_l} = 2{r_S}\], \[{L_l} = \dfrac{1}{4}{L_S}\], \[{t_l} = 25\min \] and \[{t_S} = 16\min \]

Substitute the value of in the above equation, then,
\[\dfrac{{{K_{larger}}}}{{{K_{smaller}}}} = \left( {\dfrac{1}{4}} \right) \times {\left( {\dfrac{1}{2}} \right)^2} \times \dfrac{{16}}{{25}} \\ \]
\[\Rightarrow \dfrac{{{K_{larger}}}}{{{K_{smaller}}}} = \left( {\dfrac{1}{4}} \right) \times \left( {\dfrac{1}{4}} \right) \times \dfrac{{16}}{{25}} \\ \]
\[\Rightarrow \dfrac{{{K_{larger}}}}{{{K_{smaller}}}} = \dfrac{1}{{16}} \times \dfrac{{16}}{{25}} \\ \]
\[\Rightarrow \dfrac{{{K_{larger}}}}{{{K_{smaller}}}} = \dfrac{1}{{25}}\]
That is,
\[{K_{larger}}:{K_{smaller}} = 1:25\]
Therefore, the ratio of the thermal conductivity is 1:25.

Hence, option A is the correct answer.

Note: Do not get confused with the formula for thermal conductivity and the rate of heat transfer, these two are related to each other. The thermal conductivity of a material depends on its temperature, density of material etc.