Question

A mercury thermometer has a glass bulb of interior volume 0.100 cm3 at 10°c. the glass capillary 10) tube above the bulb has an inner cross-sectional area of 0.012 mm2. the coefficient of volume expansion of mercury is 1.8 × 10-4 k-1. if the expansion of the glass is negligible, how much will the mercury rise in the capillary tube when the temperature rises from 5°c to 35°c if the bulb was full at 5°c?

Answer 1

Initial volume of mercury is
V = 0.1 cm³

The temperature rise is 35 - 5 = 30 ⁰C = 30 ⁰K.

Because the coefficient of volume expansion is 1.8x10⁻⁴ 1/K, the change in volume of the mercury is 
ΔV = (1.8x10⁻⁴ 1/K)*(30 ⁰K)(0.1 cm³) = 5.4x10⁻⁴ cm³

The cross sectional area of the tube is
A = 0.012 mm² = (0.012x10⁻² cm²).
Therefore the rise of mercury in the tube is
h = ΔV/A
   = (5.4x10⁻⁴ cm³)/(0.012x10⁻² cm²)
   = 4.5 cm

Answer: 4.5 cm

Answer 2

The mercury will rise 4.5 cm in the capillary tube.

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Further explanation

This problem is about Density.

Density is the ratio of mass to the volume of the object.

$\large {\boxed {\rho = \frac{ m }{ V } } }$

ρ = density of object ( kg / m³ )

m = mass of object ( kg )

V = volume of object ( m³ )

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The volume expansion of object could be calculated using this following formula.

$\boxed {V = Vo ( 1 + 3 \alpha \Delta t )}$

where:

V = final volume after expansion ( m³ )

Vo = initial volume before expansion ( m³ )

α = coefficient of expansion ( /K )

Δt = change in temperature ( K )

Let's tackle the problem!

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Given:

initial volume = Vo = 0.100 cm³

cross-sectional area = A = 0.012 mm² = 0.012 × 10⁻² cm²

coefficient of volume expansion of mercury = γ = 1.8 × 10⁻⁴ /K

change in temperature = Δt = 35 - 5 = 30°C

Asked:

the rise of mercury column = h = ?

Solution:

$h = \Delta V \div A$

$h = (V_o \gamma \Delta t) \div A$

$h = (0.100 \times 1.8 \times 10^{-4} \times 30) \div (0.012 \times 10^{-2})$

$h = 4.5 \texttt{ cm}$

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Answer details

Grade: High School

Subject: Physics

Chapter: Expansion