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Question: A copper cup of 100 \[c{m^3}\] capacity is fully filled with glycerine at \[{25^o}C\] . What is the ...

A copper cup of 100 cm3c{m^3} capacity is fully filled with glycerine at 25oC{25^o}C . What is the required temperature to spill out the 2.52 cm3c{m^3} of glycerine from the cup? (The coefficient of volume expansion of glycerine is 5.1×10−4/Co5.1 \times {10^{ - 4}}/{C^o} and the coefficient of linear expansion of copper is 17×10−6/Co17 \times {10^{ - 6}}/{C^o} ).

Explanation

Solution

Here we have to use the concept of volume expansion and the relation of volume expansion to the coefficient of volume expansion denoted by β\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:
⇒ΔVgV0=β×ΔT\Rightarrow \dfrac{{\Delta {V_g}}}{{{V_0}}} = \beta \times \Delta T
Where ΔVg\Delta {V_g} is the change in volume of glycerine due to heating
V0{V_0} is the original volume
β\beta is the coefficient of volume expansion
ΔT\Delta T is the change in temperature ( T2−T1{T_2} - {T_1} )
We are given all the data except the value of T2{T_2} .
Thus we can substitute this and get the following equation:
⇒ΔVg=V0×β×ΔT\Rightarrow \Delta {V_g} = {V_0} \times \beta \times \Delta T
⇒ΔVg=100×5.1×10−4×(T2−25)\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:
⇒ΔVcV0=3×α×ΔT\Rightarrow \dfrac{{\Delta {V_c}}}{{{V_0}}} = 3 \times \alpha \times \Delta T
Here ΔVc\Delta {V_c} is the change in volume of copper.
α\alpha is the coefficient of linear expansion of copper and is given in the question as 17×10−6/Co17 \times {10^{ - 6}}/{C^o} .
The equation to find the change in volume due to linear expansion is given by:
⇒ΔVc=V0×3×α×ΔT\Rightarrow \Delta {V_c} = {V_0} \times 3 \times \alpha \times \Delta T
Substituting the values we get.

⇒ΔVc=100×3×17×10−6×(T2−25) \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.
⇒ΔV=ΔVg−ΔVc\Rightarrow \Delta V = \Delta {V_g} - \Delta {V_c}
⇒ΔV=100×5.1×10−4×(T2−25)−[100×3×17×10−6×(T2−25)]\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)]
⇒ΔV=100×(T2−25)[5.1×10−4−3×17×10−6]\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 cm3c{m^3}
⇒2.52=100×(T2−25)[5.1×10−4−3×17×10−6]\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,
⇒2.52=0.046×(T2−25)\Rightarrow 2.52 = 0.046 \times ({T_2} - 25)
⇒54.9=(T2−25)\Rightarrow 54.9 = ({T_2} - 25)
By solving for T2{T_2} ,
⇒T2=54.9+25\Rightarrow {T_2} = 54.9 + 25
⇒T2=79.9oc\Rightarrow {T_2} = {79.9^o}c
Thus the required temperature to spill out the 2.52 cm3c{m^3} of glycerine from the cup will be 79.9oc{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.