Question

A copper cup of 100 \[c{m^3}\] capacity is fully filled with glycerine at \[{25^o}C\] . What is the required temperature to spill out the 2.52 \[c{m^3}\] of glycerine from the cup? (The coefficient of volume expansion of glycerine is \[5.1 \times {10^{ - 4}}/{C^o}\] and the coefficient of linear expansion of copper is \[17 \times {10^{ - 6}}/{C^o}\] ).

Answer 1

Hint: Here we have to use the concept of volume expansion and the relation of volume expansion to the coefficient of volume expansion denoted by \[\beta \] . Volume expansion represents the change in volume due to thermal expansion and is defined as the ratio of the change in volume to the original volume.

Complete answer:
We know that volume expansion can be represented by the following formula:
\[ \Rightarrow \dfrac{{\Delta {V_g}}}{{{V_0}}} = \beta \times \Delta T\]
Where \[\Delta {V_g}\] is the change in volume of glycerine due to heating
\[{V_0}\] is the original volume
\[\beta \] is the coefficient of volume expansion
\[\Delta T\] is the change in temperature ( \[{T_2} - {T_1}\] )
We are given all the data except the value of \[{T_2}\] .
Thus we can substitute this and get the following equation:
\[ \Rightarrow \Delta {V_g} = {V_0} \times \beta \times \Delta T\]
\[ \Rightarrow \Delta {V_g} = 100 \times 5.1 \times {10^{ - 4}} \times ({T_2} - 25)\]
Now for finding the change in the volume of copper we can use the following formula:
\[ \Rightarrow \dfrac{{\Delta {V_c}}}{{{V_0}}} = 3 \times \alpha \times \Delta T\]
Here \[\Delta {V_c}\] is the change in volume of copper.
\[\alpha \] is the coefficient of linear expansion of copper and is given in the question as \[17 \times {10^{ - 6}}/{C^o}\] .
The equation to find the change in volume due to linear expansion is given by:
\[ \Rightarrow \Delta {V_c} = {V_0} \times 3 \times \alpha \times \Delta T\]
Substituting the values we get.

\[ \Rightarrow \Delta {V_c} = 100 \times 3 \times 17 \times {10^{ - 6}} \times ({T_2} - 25)\]
Now the total change in volume of the system is given by the difference between the change in volume of glycerine due to volume expansion and the change in volume of copper due to linear expansion.
\[ \Rightarrow \Delta V = \Delta {V_g} - \Delta {V_c}\]
\[ \Rightarrow \Delta V = 100 \times 5.1 \times {10^{ - 4}} \times ({T_2} - 25) - [100 \times 3 \times 17 \times {10^{ - 6}} \times ({T_2} - 25)]\]
\[ \Rightarrow \Delta V = 100 \times ({T_2} - 25)[5.1 \times {10^{ - 4}} - 3 \times 17 \times {10^{ - 6}}]\]
We are given that the Increase in volume of glycerine is 2.52 \[c{m^3}\]
\[ \Rightarrow 2.52 = 100 \times ({T_2} - 25)[5.1 \times {10^{ - 4}} - 3 \times 17 \times {10^{ - 6}}]\]
We can simplify the RHS and obtain,
\[ \Rightarrow 2.52 = 0.046 \times ({T_2} - 25)\]
\[ \Rightarrow 54.9 = ({T_2} - 25)\]
By solving for \[{T_2}\] ,
\[ \Rightarrow {T_2} = 54.9 + 25\]
\[ \Rightarrow {T_2} = {79.9^o}c\]
Thus the required temperature to spill out the 2.52 \[c{m^3}\] of glycerine from the cup will be \[{79.9^o}c\].

Note:
Thermal expansions can be of three different types. Linear, Area, and volume expansion and all three have different formulas. In the case of liquids like the one mentioned in this question, there is always volume expansion. In the case of rods, there is the linear expansion and in the case of sheets, there will be area expansion.