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Question: The density of mercury is \(13.6gc{{m}^{-3}}\)at \[{{0}^{\circ }}\text{C}\] and its coefficient of c...

The density of mercury is 13.6gcm−313.6gc{{m}^{-3}}at 0∘C{{0}^{\circ }}\text{C} and its coefficient of cubical expansion is 1.82×10−4C−11.82\times {{10}^{-4}}{{C}^{-1}}. Calculate the density of mercury at 50∘C\text{5}{{\text{0}}^{\circ }}C.

Explanation

Solution

- Hint: Density of mercury can be taken as the mass, if we are using unit volume. Change in volume is directly proportional to the change in temperature. We can find the density of mercury from the new volume since the mass of the mercury is not varying.

Complete step-by-step solution -

Mass of the mercury is given, m=13.6gm=13.6g
Coefficient of cubical expansion is given, α=1.82×10−4C−1\alpha =1.82\times {{10}^{-4}}{{C}^{-1}}
Change in temperature, ΔT=50\Delta T=50
Let consider the mercury have volume (V) of 1 cm3cm^3
When mercury is heated to 50∘C{{50}^{\circ }}\text{C}, volume will change due to the expansion.
Change in volume, ΔV=VαΔT\Delta V=V\alpha \Delta T
Assign the values into the equation, ΔV=1×1.82×10−4×50\Delta V=1\times 1.82\times {{10}^{-4}}\times 50
Then, ΔV=9.1×10−3cm3\Delta V=9.1\times {{10}^{-3}}c{{m}^{3}}
New volume of the mercury (Vm)\text{(}{{\text{V}}_{m}}) after the expansion will be,
Vm=V+ ΔV{{V}_{m}}=V+~\Delta V
Assign the values to the equation,
Vm=1+9.1×10−3cm3{{V}_{m}}=1+9.1\times {{10}^{-3}}c{{m}^{3}}
Vm=1.0091 cm3{{V}_{m}}=1.0091~c{{m}^{3}}
From this volume, we can find the density (ρ)(\rho ) of the mercury at a given temperature.
ρ=mVm\rho =\dfrac{m}{{{V}_{m}}}
After assigning the values to the equation, ρ=13.61.0091\rho =\dfrac{13.6}{1.0091}
Density of the mercury, ρ=13.48 gcm−3\rho =13.48~gc{{m}^{-3}}

Additional information: Coefficient of cubical expansion is the increment in the volume of unit volume of any state for a rise of temperature of 1∘C{{1}^{\circ }}\text{C} at constant pressure. It is also known as coefficient of expansion or coefficient of thermal expansion.
Invar is a special alloy of iron and nickel, which has a coefficient of expansion nearly zero. The name of that alloy came from ‘Invariable’.

Note: Before the calculation we need to assign a unit volume for the mercury as V. Then only we can equate this density as the mass of the mercury. Changes in volume have a direct relationship with this volume. The volume change will happen due to the temperature rise, but it is not showing directly in the question. However, we have to think like that, because the question already provided the coefficient of cubical expansion of mercury. It is advised to learn the relation of change in volume with the coefficient of cubical expansion.