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Ainat [17]
3 years ago
7

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?

Physics
2 answers:
Nadya [2.5K]3 years ago
7 0
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
liraira [26]3 years ago
6 0

The mercury will rise 4.5 cm in the capillary tube.

\texttt{ }

<h3>Further explanation</h3>

This problem is about Density.

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

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

<em>ρ = density of object ( kg / m³ )</em>

<em>m = mass of object ( kg )</em>

<em>V = volume of object ( m³ )</em>

\texttt{ }

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

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

where:

<em>V = final volume after expansion ( m³ )</em>

<em>Vo = initial volume before expansion ( m³ )</em>

<em>α = coefficient of expansion ( /K )</em>

<em>Δt = change in temperature ( K )</em>

<em>Let's tackle the problem!</em>

\texttt{ }

<u>Given:</u>

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

<u>Asked:</u>

the rise of mercury column = h = ?

<u>Solution:</u>

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}

\texttt{ }

<h3>Learn more</h3>
  • Velocity of Runner : brainly.com/question/3813437
  • Kinetic Energy : brainly.com/question/692781
  • Acceleration : brainly.com/question/2283922
  • The Speed of Car : brainly.com/question/568302

\texttt{ }

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Expansion

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6 0
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