Question:

Two spheres 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 complete melting of ice in the larger sphere is $25$ minute and for smaller one is $16$ minute, the ratio of thermal conductivities of the materials of larger spheres to that of smaller sphere is

Updated On: Jun 18, 2022
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The Correct Option is D

Solution and Explanation

Radius of small sphere $=r$
Thickness of small sphere $=t$
Radius of bigger sphere $=2 r$
Thickness of bigger sphere $=t / 4$
Mass of ice melted $=($ volume of sphere $) \times$ (density of ice)
Let $K_{1}$ and $K_{2}$ be the thermal conductivities of larger and smaller sphere.
For bigger sphere,
$\frac{K_{1} 4 \pi(2 r)^{2} \times 100}{t /4}=\frac{\frac{4}{3} \pi(2 r)^{3} \rho L}{25 \times 60}$
For smaller sphere,
$\frac{K_{2} \times 4 \pi r^{2} \times 100}{t}=\frac{\frac{4}{3} \pi r^{3} \rho L}{16 \times 60}$
$\therefore \frac{K_{1}}{K_{2}}=\frac{8}{25}$
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Concepts Used:

Newton’s Law of Cooling

Newton’s law of cooling states that the rate of heat loss from a body is directly proportional to the difference in temperature between the body and its surroundings. 

Derivation of Newton’s Law of Cooling

Let a body of mass m, with specific heat capacity s, is at temperature T2 and T1 is the temperature of the surroundings. 

If the temperature falls by a small amount dT2 in time dt, then the amount of heat lost is,

dQ = ms dT2

The rate of loss of heat is given by,

dQ/dt = ms (dT2/dt)                                                                                                                                                                              ……..(2)

Compare the equations (1) and (2) as,

– ms (dT2/dt) = k (T2 – T1)

Rearrange the above equation as:

dT2/(T2–T1) = – (k / ms) dt

dT2 /(T2 – T1) = – Kdt 

where K = k/m s

Integrating the above expression as,

loge (T2 – T1) = – K t + c

or 

T2 = T1 + C’ e–Kt

where C’ = ec