Question

An antimony ball of volume $15{\text{cm}}^3$ is heated to $210{}^{\circ }\text{C}$ which caused the volume to become $15.045{\text{cm}}^3$, if $\gamma$ for antimony is $15\times {10}^{-6}{}^{\circ }{\text{C}}^{-1}$, what is the initial temperature of the ball?

• A. 10

Answer 1

The correct option is A 10
Given:
The coefficient of cubical thermal expansion for the antimony ball, ${\gamma }_{\text{Sb}}=15\times {10}^{-6}{}^{\circ }{\text{C}}^{-1}$
The final temperature of the ball, ${\text{T}}_{\text{f}}=210{}^{\circ }\text{C}$
Let the initial temperature be ${\text{T}}_{\text{i}}{}^{\circ }\text{C}$
Change in the volume, $\Delta \text{V}=15.045-15=0.045{\text{cm}}^3$
We know that
$\gamma =\frac{\Delta \text{V}}{{\text{V}}_0\times \Delta \text{T}}$, where ${\text{V}}_0$ is the initial volume, and $\Delta \text{T}$ is the rise in the temperature $={\text{T}}_{\text{f}}-{\text{T}}_{\text{i}}$

$⟹\Delta \text{T}=\frac{\Delta \text{V}}{{\text{V}}_0\gamma }=\frac{0.045}{15\times 15\times {10}^{-6}}$

$⟹\Delta \text{T}=200{}^{\circ }\text{C}$
$⟹{\text{T}}_{\text{f}}-{\text{T}}_{\text{i}}=210-{\text{T}}_{\text{i}}=200{}^{\circ }\text{C}$
$⟹{\text{T}}_{\text{i}}=210-200=10{}^{\circ }\text{C}$

$\therefore$ The initial temperature of the ball was $10{}^{\circ }\text{C}$.