Question

A $\mathbf{20}\mathbf{.0}\mathbf{\text{g}}$copper ring at$\mathbf{0}\mathbf{.000}\mathbf{\text{°C}}$has an inner diameter of.$\mathit{D}\mathbf{=}\mathbf{}\mathbf{2}\mathbf{.54000}\mathbf{\text{cm}}$An aluminum sphere at$\mathbf{100}\mathbf{.0}\mathbf{\text{°C}}$has a diameter of$\mathit{d}\mathbf{=}\mathbf{2}\mathbf{.545}\mathbf{}\mathbf{08}\mathbf{\text{cm}}$. The sphere is put on top of the ring (Figure), and the two are allowed to come to thermal equilibrium, with no heat lost to the surroundings. The sphere just passes through the ring at the equilibrium temperature. What is the mass of the sphere?

Short answer

The mass of the sphere is$8.71\times {10}^{-3}\text{kg}$

Step-by-step solution

Identification of given data

i) The diameter of the ring at$0°\text{C}$ is$\left(D_{r0}\right)=2.54000\text{\hspace{0.17em}cm}$

ii) The diameter of the ring at$100°\text{C}$ is$\left(D_{r0}\right)=2.54508\text{\hspace{0.17em}cm}$

iii) Mass of ring,$\left(M_r\right)=0.0200\text{\hspace{0.17em}kg.}$

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

v) Coefficient of linear expansion of aluminum,$\left({\alpha }_a\right)=23\times {10}^{-6}\text{/°C}$

vi) Coefficient of linear expansion of copper,$\left({\alpha }_c\right)=17\times {10}^{-6}/°\text{C}$

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

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

Significance of specific heat

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$\mathbf{0}\mathbf{\text{°C}}$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$\mathbf{}\mathit{D}\mathbf{=}{\mathit{D}}_{\mathbf{r}\mathbf{0}}\mathbf{1}\mathbf{+}{\mathit{\alpha }}_{\mathbf{c}}{\mathit{T}}_{\mathbf{f}}$ and$\mathit{D}\mathbf{=}{\mathit{D}}_{\mathbf{s}\mathbf{0}}\mathbf{[}\mathbf{1}\mathbf{+}{\mathit{\alpha }}_{\mathbf{a}}\mathbf{(}{\mathbf{T}}_{\mathbf{f}}\mathbf{-}{\mathbf{T}}_{\mathbf{i}}\mathbf{)}\mathbf{]}$is given as:

${\mathit{T}}_{\mathbf{f}}\mathbf{=}\frac{{\mathbf{D}}_{\mathbf{r}\mathbf{0}}\mathbf{-}{\mathbf{D}}_{\mathbf{s}\mathbf{0}}\mathbf{+}{\mathbf{D}}_{\mathbf{s}\mathbf{0}}{\mathbf{\alpha }}_{\mathbf{a}}{\mathbf{T}}_{\mathbf{i}}}{{\mathbf{D}}_{\mathbf{s}\mathbf{0}}{\mathbf{\alpha }}_{\mathbf{a}}\mathbf{-}{\mathbf{D}}_{\mathbf{r}\mathbf{0}}{\mathbf{\alpha }}_{\mathbf{c}}}$

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, ${\mathit{T}}_{\mathbf{f}}\mathbf{=}\frac{{\mathbf{D}}_{\mathbf{r}\mathbf{0}}\mathbf{-}{\mathbf{D}}_{\mathbf{s}\mathbf{0}}\mathbf{+}{\mathbf{D}}_{\mathbf{s}\mathbf{0}}{\mathbf{\alpha }}_{\mathbf{a}}{\mathbf{T}}_{\mathbf{i}}}{{\mathbf{D}}_{\mathbf{s}\mathbf{0}}{\mathbf{\alpha }}_{\mathbf{a}}\mathbf{-}{\mathbf{D}}_{\mathbf{r}\mathbf{0}}{\mathbf{\alpha }}_{\mathbf{c}}}$ …(i)

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, ${\alpha }_a$iscoefficient of linear expansion of aluminum, $T_i$is the initial temperature,${\alpha }_c$is the coefficient of linear expansion of copper.

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

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.

Determining the mass of the sphere

If the ring diameter at${0.000}^{0}C$is$D_{r0}$, then its diameter when ring and sphere are in thermal equilibrium is$D=D_{r0}1+{\alpha }_c{T}_f$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$D=D_{s0}[1+{\alpha }_a(T_f-T_i)]$.

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

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

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

$\begin{array}{c}{M}_s=\frac{(386\text{J/kg}\cdot \text{K})\times \left(0.0200\text{kg}\right)\times (50.38°\text{C})}{(900\text{J/kg}\cdot \text{K})(100°\text{C}-50.38°\text{C})}\\ =\frac{388.9336}{44658}\text{kg}\\ =0.008709\text{\hspace{0.17em}kg}\\ =8.71\times {10}^{-3}\text{\hspace{0.17em}kg}\end{array}$

Hence, the value of the mass of the sphere is .$8.71\times {10}^{-3}\text{\hspace{0.17em}kg}$