Question

A spherical brass shell has an interior volume of 1.26 x 10-3 m3. Within this interior volume is a solid steel ball with volume of 0.63 x 10-3 m3. The space between the steel ball and the inner surface of the brass shell's filled completely with mercury. A hole's drilled through the brass, and the temperature of the arrangement in increased by 18 C°. How much mercury spills out?

Answer 1

Answer:

ΔV = 2.04 x 10⁻⁶ m³ = 2.04 cm³

Explanation:

First we will find the initial volume of mercury by subtracting the volume of the steel ball from the interior volume of the brass shell:

V = Vs - Vb

where,

V = Initial Volume of Mercury  = ?

Vb = Volume of brass shell = 1.26 x 10⁻³ m³

Vs = Volume of Steel Ball = 0.63  x 10⁻³ m³

Therefore,

V = (1.26 x 10⁻³ m³) - (0.63 x 10⁻³ m³)

V = 0.63 x 10⁻³ m³

Now, we find the change in volume of mercury that will spill out due to increase in temperature:

ΔV = βVΔT

where,

ΔV = Amount of Mercury Spilled out = ?

β = Coefficient of Volumetric Expansion of mercury = 1.8 x 10⁻⁴ °C⁻¹

ΔT = Increase in temperature = 18 °C

Therefore,

ΔV = (1.8 x 10⁻⁴ °C⁻¹)(0.63 x 10⁻³ m³)(18 °C)

ΔV = 2.04 x 10⁻⁶ m³ = 2.04 cm³

Answer 2

The volume of mercury that spills out due to increase in temperature is $2.04 \times 10^{-6} \ m^3$.

The given parameters;

• outer volume, V₁ = 1.26 x 10⁻³ m³
• inner volume, V₂ = 0.63 x 10⁻³ m³
• change in temperature, Δt = 18 ⁰C

The volume of mercury in the given shell is calculated as;

$V= V_1 - V_2\\\\V= 1.26 \times 10^{-3} \ - \ 0.63 \times 10^{-3} \\\\V = 6.3 \times 10^{-4} \ m^3$

The change in the volume of the mercury is calculated as follows;

$\Delta V = \beta V \Delta t$

where;

• β is the coefficient of volumetric expansion of mercury

$\Delta V = (1.8 \times 10^{-4} ) \times (6.3\times 10^{-4}) \times 18\\\\\Delta V = 2.04 \times 10^{-6} \ m^3$

Thus, the volume of mercury that spills out due to increase in temperature is $2.04 \times 10^{-6} \ m^3$.

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