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Question: Calculate the coefficient of cubic expansion of a zinc bar. Whose volume has increased 0.25 \(m^3 \)...

Calculate the coefficient of cubic expansion of a zinc bar. Whose volume has increased 0.25 m3m^3 from 0.3 m3m^3 due to change in its temperature of 50 K.

Explanation

Solution

When a solid of volume V is heated, the volume increases by the proportion that depends on the coefficient of volume expansion and also on the rise in temperature that it has gone through upon action of heat on the solid.
Formula Used:
The coefficient of thermal expansion is expressed as:
γ=VVV×ΔT\gamma = {{V' - V} \over {V \times \Delta T}}.
V’ is the increased (or decreased) volume, ΔT\Delta T is the change in the temperature.

Complete answer:
The coefficient of thermal volume expansion is defined as the increase in volume per unit volume per degree rise in temperature. The volume expansion, when temperature is increased byΔT\Delta T , is given as:
V=V(1+γΔT)V' = V(1 + \gamma \Delta T).
Here, we can clearly see that the original volume is V and a fractional increase in volume gets added, according to the temperature and coefficient of volume expansion to the original volume to result in the increased total volume.
Manipulating the terms in the above expression, we can easily find the formula for γ\gamma
as:
γ=VVV×ΔT\gamma = {{V' - V} \over {V \times \Delta T}}
From the question, we can write V = 0.3 m3m^3 , to which we add 0.25, to get V’ = 0.55m3m^3 but still, V’-V is directly substituted 0.25 m3m^3
and the change in the temperature, ΔT\Delta T
is given as 50K.
Substituting these values, we find:
γ=0.25(0.3)×(50)=0.1667K1\gamma = {{0.25} \over {(0.3) \times (50)}} = 0.1667 K^{ - 1}
Therefore for the zinc bar, the coefficient of volume expansion comes as 0.1667K1K^{ - 1} .

Note:
The expansion coefficient in this case is for solids and not gases. In gases it is inversely proportional to absolute temperature. The temperature substituted here is in Kelvins. One has to be careful about temperature units in such types of questions. The coefficient here is not unitless here. So, one must write the units by properly writing the temperature units.