91Ó°ÊÓ

i) The diameter of the ring at$0°\text{C}$ is$\left(D_{r0}\right)=2.54000\text{ c³¾}$

ii) The diameter of the ring at$100°\text{C}$ is$\left(D_{r0}\right)=2.54508\text{ c³¾}$

iii) Mass of ring,$\left(M_r\right)=0.0200\text{ k²µ.}$

iv) Initial temperature,$\left(T_i\right)100°\text{C}$

v) Coefficient of linear expansion of aluminum,$\left({\mathrm{Î\pm }}_a\right)=23×{10}^{−6}\text{/°°ä}$

vi) Coefficient of linear expansion of copper,$\left({\mathrm{Î\pm }}_c\right)=17×{10}^{−6}/°\text{C}$

vii) Specific heat of copper,$\left(C_c\right)=386\text{J/kg}â‹\dots \text{K}$

viiii) Specific heat of aluminum,$\left(C_a\right)=900\text{J/kg}â‹\dots \text{K}$

The specific heat capacity is the amount of heat required to raise the temperature of unit mass of a substance by one degree.

We are given that the diameter of the ring at$\mathbf{0}\mathbf{\text{°°ä}}$isand ${\mathit{D}}_{\mathbf{r}\mathbf{0}}$the diameter of the sphere at initial temperature is${\mathit{D}}_{\mathbf{s}\mathbf{0}}$. So, we can find first the final temperature by using two equations$\mathbf{}\mathit{D}\mathbf{=}{\mathit{D}}_{\mathbf{r}\mathbf{0}}\mathbf{1}\mathbf{+}{\mathit{Î\pm }}_{\mathbf{c}}{\mathit{T}}_{\mathbf{f}}$ and$\mathit{D}\mathbf{=}{\mathit{D}}_{\mathbf{s}\mathbf{0}}\mathbf{[}\mathbf{1}\mathbf{+}{\mathit{Î\pm }}_{\mathbf{a}}\mathbf{(}{\mathbf{T}}_{\mathbf{f}}\mathbf{−}{\mathbf{T}}_{\mathbf{i}}\mathbf{)}\mathbf{]}$is given as:

${\mathit{T}}_{\mathbf{f}}\mathbf{=}\frac{{\mathbf{D}}_{\mathbf{r}\mathbf{0}}\mathbf{−}{\mathbf{D}}_{\mathbf{s}\mathbf{0}}\mathbf{+}{\mathbf{D}}_{\mathbf{s}\mathbf{0}}{\mathbf{Î\pm }}_{\mathbf{a}}{\mathbf{T}}_{\mathbf{i}}}{{\mathbf{D}}_{\mathbf{s}\mathbf{0}}{\mathbf{Î\pm }}_{\mathbf{a}}\mathbf{−}{\mathbf{D}}_{\mathbf{r}\mathbf{0}}{\mathbf{Î\pm }}_{\mathbf{c}}}$

It is given that the sphere and the ring are at thermal equilibrium.

At equilibrium, the heat absorbed by the ring = the heat released by the sphere,so we write

${\mathit{C}}_{\mathbf{c}}{\mathit{M}}_{\mathbf{r}}{\mathit{T}}_{\mathbf{f}}\mathbf{=}{\mathit{C}}_{\mathbf{a}}{\mathit{M}}_{\mathbf{s}}\mathbf{(}{\mathbf{T}}_{\mathbf{i}}\mathbf{−}{\mathbf{T}}_{\mathbf{f}}\mathbf{)}$

From this equation, we find the mass of sphere as

${\mathit{M}}_{\mathbf{s}}\mathbf{=}\frac{{\mathbf{C}}_{\mathbf{c}}{\mathbf{M}}_{\mathbf{r}}{\mathbf{T}}_{\mathbf{f}}}{{\mathbf{C}}_{\mathbf{a}}\mathbf{(}{\mathbf{T}}_{\mathbf{i}}\mathbf{−}{\mathbf{T}}_{\mathbf{f}}\mathbf{)}}$

Formula:

The final temperature of the system, ${\mathit{T}}_{\mathbf{f}}\mathbf{=}\frac{{\mathbf{D}}_{\mathbf{r}\mathbf{0}}\mathbf{−}{\mathbf{D}}_{\mathbf{s}\mathbf{0}}\mathbf{+}{\mathbf{D}}_{\mathbf{s}\mathbf{0}}{\mathbf{Î\pm }}_{\mathbf{a}}{\mathbf{T}}_{\mathbf{i}}}{{\mathbf{D}}_{\mathbf{s}\mathbf{0}}{\mathbf{Î\pm }}_{\mathbf{a}}\mathbf{−}{\mathbf{D}}_{\mathbf{r}\mathbf{0}}{\mathbf{Î\pm }}_{\mathbf{c}}}$ …(¾±)

Here,$T_f$ is the final temperature,$D_{r0}$ is the initial diameter of the ring,$D_{s0}$ is the initial diameter of the sphere, ${\mathrm{Î\pm }}_a$iscoefficient of linear expansion of aluminum, $T_i$is the initial temperature,${\mathrm{Î\pm }}_c$is the coefficient of linear expansion of copper.

The mass of the sphere is given as: $M_s=\frac{C_c{M}_r{T}_f}{C_a(T_i−T_f)}$ …(¾±¾±)

Here,$M_s$is mass of sphere,data-custom-editor="chemistry" $C_c$ is specific heat capacity of copper, $C_a$is specific heat capacity of aluminum.

If the ring diameter at${0.000}^{0}C$is$D_{r0}$, then its diameter when ring and sphere are in thermal equilibrium is$D=D_{r0}1+{\mathrm{Î\pm }}_c{T}_f$where$T_f$is the final temperature andis the coefficient of linear expansion of copper.

Similarly, if the sphere diameter at$(T_i={100}^{0}C)$is$D_{s0}$, then its diameter at its final temperature is$D=D_{s0}[1+{\mathrm{Î\pm }}_a(T_f−T_i)]$.

Then the final temperature using equation (i) is given as:

$\begin{array}{c}{T}_f=\frac{2.54000\text{cm}−2.54508\text{cm}+\left(2.54508\text{cm}\right)×(23×{10}^{−6}/°\text{C})×100/°\text{C}}{\left(2.54508\text{cm}\right)×(23×{10}^{−6}/°\text{C})−\left(2.54000\text{cm}\right)×(17×{10}^{−6}/°\text{C})}\\ =\frac{773.684}{15.3734}°\text{C}\\ =50.38\text{ }°\text{C}\end{array}$

Then, substituting these values in equation (ii), we can get the mass of the sphere as:

$\begin{array}{c}{M}_s=\frac{(386\text{J/kg}â‹\dots \text{K})×\left(0.0200\text{kg}\right)×(50.38°\text{C})}{(900\text{J/kg}â‹\dots \text{K})(100°\text{C}−50.38°\text{C})}\\ =\frac{388.9336}{44658}\text{kg}\\ =0.008709\text{ k²µ}\\ =8.71×{10}^{−3}\text{ k²µ}\end{array}$

Hence, the value of the mass of the sphere is .$8.71×{10}^{−3}\text{ k²µ}$