A solid aluminium sphere and a solid copper sphere of twice the radius are heated to the same temperature and are allowed to cool under identical surrounding temperatures. Assume that the emissivity of both the spheres is the same. Find the ratio of: 1. The rate of heat loss from the aluminium sphere to the rate of heat loss from the copper sphere. 2. The rate of fall of temperature of the aluminium sphere to the rate of fall of temperature of the copper sphere. The specific heat capacity of aluminium = 900Jkg^ -1^ C^ -1 and that of copper = 390Jkg^ -1^ C^ -1 . The density of copper = 3.4 times the density of aluminium.

E Energy radiated per unit area per unit time Rate of heat flow Energy radiated 1. Per time =E So, E_Al= e ^4 e ^4 = 4 ^2 4 (2r)^2 =1 4 [ 1:4 ] 2. Emissivity of both are same = m_1S_1dT_1 m_2S_2dT_2 =1 dT_1 dT_2 = m_2S_2 m_1S_1 = s_14 _1^3 _2 s_24 _2^3 _1 =1 900 3.4 8 390 =1:2:9

A solid aluminium sphere and a solid copper sphere of twice the r…

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Question

A solid aluminium sphere and a solid copper sphere of twice the radius are heated to the same temperature and are allowed to cool under identical surrounding temperatures. Assume that the emissivity of both the spheres is the same. Find the ratio of:

The rate of heat loss from the aluminium sphere to the rate of heat loss from the copper sphere.
The rate of fall of temperature of the aluminium sphere to the rate of fall of temperature of the copper sphere. The specific heat capacity of aluminium $= 900Jkg^{-1^\circ }C^{-1}$ and that of copper $= 390Jkg^{-1^\circ }C^{-1}$. The density of copper $= 3.4$ times the density of aluminium.

✓

Answer

$E \rightarrow $ Energy radiated per unit area per unit time Rate of heat flow $\rightarrow $ Energy radiated

Per time $=\text{E}\times\text{A}$

So, $\text{E}_\text{Al}=\frac{\text{e}\sigma\text{T}^4\times\text{A}}{\text{e}\sigma\text{T}^4\times\text{A}}$
$=\frac{4\pi\text{r}^2}{4\pi(2\text{r})^2}$
$=\frac{1}{4}$ $\big[\therefore1:4\big]$

Emissivity of both are same

$=\frac{\text{m}_1\text{S}_1\text{dT}_1}{\text{m}_2\text{S}_2\text{dT}_2}=1$
$\Rightarrow\frac{\text{dT}_1}{\text{dT}_2}=\frac{\text{m}_2\text{S}_2}{\text{m}_1\text{S}_1}$
$=\frac{\text{s}_14\pi\text{r}_1^3\times\text{S}_2}{\text{s}_24\pi\text{r}_2^3\times\text{S}_1}$
$=\frac{1\times\pi\times900}{3.4\times8\pi\times390}$
$=1:2:9$

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