First slide

Heat transfer-Radiation

Question

A solid copper cube and a solid copper sphere, both of same mass & emissivity are heated to same initial temperature and kept under identical conditions. What is the ratio of their initial rate of fall of temperature?

Moderate

A

${(\frac{6}{π})}^{\frac{1}{3}}$
B

${(\frac{6}{π})}^{\frac{2}{3}}$
C

${(\frac{6}{π})}^{3}$
D

$\frac{6}{π}$

Solution

If $' a'$ is the length of cube, surface area $S_{cube} = 6 a^{2}$
If $' r'$ is the radius of sphere, surface area $S_{sphere} = 4 π r^{2}$
Both the cube and sphere have the same volume as given
Volume $V = a^{3}$ $(∵ V = \frac{4}{3} π r^{3})$
$\begin{array}{l} ∴ a^{3} = \frac{4}{3} π r^{3} \\ \Rightarrow \frac{a}{r} = {(\frac{4 π}{3})}^{\frac{1}{3}} \\ \frac{{Area}_{cube}}{{Area}_{sphere}} = \frac{6 a^{2}}{4 π r^{2}} \\ = \frac{6}{4 π} {(\frac{a}{r})}^{2} = \frac{6}{4 π} \times {(\frac{4 π}{3})}^{\frac{2}{3}} \\ = {(\frac{6}{π})}^{\frac{1}{3}} \\ ∵ mass are equal \end{array}$
Rate of fall in temperature = Surface area
$∵$ Rate of fall in temperature= ${(\frac{6}{π})}^{\frac{1}{3}}$

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