The Hot Water Challenge!

Classical Mechanics Level 2

Problem: A sphere of mass m 0 m_{0} and radius r 0 r_{0} floats on a bath of liquid. The temperature is raised, and the sphere begins to sink at a temperature t t . If the density of the liquid is ρ \rho at a temperature t 0 t_{0} ( t > t 0 t>t_{0} ), find the coefficient of cubical expansion of the liquid(which is to be considered considerably big), neglecting the expansion of the sphere.

For the problem, take:

  1. t = 1 ∘ t=1^\circ C
  2. t 0 = 0. 5 ∘ t_{0}=0.5^\circ C
  3. ρ = 1 0 2 \rho =10^{2} kg/m^3
  4. r 0 = 0.1 r_0=0.1 metre
  5. m 0 = 1 m_0=1 kg

Note that all the data given/found out is unrealistic, and the only motive is to find the answer in variables, and later substitute the values. You may use a calculator, and express your answer in the greatest-integer form(for example, if you get the answer as 4.35, you may write- 4.)

Also, note that the main motivation for this problem lies here- 'the coefficient of cubical expansion of the liquid(which is to be considered considerably big)', although the answer does not deviate much if the case is not so.

9 3 6 8

Kushal Thaman by Kushal Thaman , 18, India

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1 solution

Kushal Thaman
Nov 28, 2018

The first part of the solution is under the assumption that γ = ’small’ \gamma=\text{'small'} . ρ c u b e = m 0 4 3 π r 3 = 3 m 4 π r 3 \rho_{cube} = \frac{m_{0}}{\frac{4}{3} \pi r^{3}} = \frac{3m}{4 \pi r^{3}} . At temperature t t , when the sphere is just about to sink, ρ c u b e = ρ = 3 m 4 π r 3 \rho_{cube} = \rho = \frac{3m}{4 \pi r^{3}} .

Now, let the mass of the liquid be m m (constant throughout the problem). We know that, for the volumetric expansion of the liquid, V t = V t 0 ( 1 + γ Δ T ) V_{t} = V_{t_{0}} (1+\gamma \Delta{T}) . Rearranging this gives 1 V t 0 = 1 V t ( 1 + γ Δ T ) \frac{1}{V_{t_{0}}} = \frac{1}{V_{t}} (1+\gamma \Delta{T}) . Multiplying on both sides by the constant quantity m m , we get:

ρ t 0 = ρ t ( 1 + γ Δ T ) \rho_{t_{0}} = \rho_{t}(1+\gamma \Delta{T}) . This gives us γ = ρ t 0 − ρ t ρ t ⋅ Δ T \gamma = \frac{\rho_{t_{0}} - \rho_{t}}{\rho_{t} \cdot \Delta{T}} . Now, simply replacing ρ t \rho_{t} as ρ \rho or 3 m 4 π r 3 \frac{3m}{4 \pi r^{3}} , and Δ T \Delta{T} as t − t 0 t-t_{0} , we obtain: γ = 4 π r 3 ρ t 0 − 3 m 3 m ( t − t 0 ) \gamma = \boxed{\frac{4 \pi r^{3} \rho_{t_{0}} - 3m}{3m(t-t_{0})}} .

For γ ≠ ’small’ \gamma \neq \text{'small'} , γ = 1 V ⋅ d V d T \gamma = \frac{1}{V} \cdot \dfrac{dV}{dT}

ρ t 0 = ρ t e γ Δ T \rho_{t_{0}}=\rho_{t} e^{\gamma \Delta{T}} .

1 Δ T ⋅ ln ⁡ ρ t 0 ρ t = γ \frac{1}{\Delta{T}} \cdot \ln{\frac{\rho_{t_{0}}}{\rho_{t}}} = \gamma

γ = 1 t − t 0 ln ⁡ ρ t 0 ⋅ 4 π r 3 3 m \gamma= \boxed{\frac{1}{t-t_{0}} \ln{\frac{\rho_{t_{0}} \cdot 4 \pi r^{3}}{3m}}} . Inserting the values, we get the answer as approximately 9.66 9.66 , so the answer is 9 \boxed{\boxed{9}}

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