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?

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

The volume of mercury that spills out due to increase in temperature is .

The given parameters;

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

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

where;

Thus, the volume of mercury that spills out due to increase in temperature is .

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