Question

Two metal bars experience an equal change in volume due to an equal change in temperature. The first bar has a coefficient of expansion twice as large as the second bar. How does the original volume of the first bar compare to the original volume of the second bar

Answer 1

Answer:

The original volume of the first bar is half of the original volume of the second bar.

Explanation:

The coefficient of cubic expansivity of substances is given by;

γ = ΔV ÷ ($V_{1}$Δθ)

Given: two metal bars with equal change in volume, equal change in temperature.

Let the volume of the first metal bar be represented by $V_{1}$, and that of the second by $V_{2}$.

Since they have equal change in volume,

Δ$V_{1}$  = Δ$V_{2}$ = ΔV

For the first metal bar,

2γ = ΔV ÷ ($V_{1}$Δθ)

⇒    Δθ =  ΔV ÷ (2γ$V_{1}$)

For the second metal bar,

γ = ΔV ÷ ($V_{2}$Δθ)

⇒  Δθ = ΔV ÷ ($V_{2}$γ)

Since they have equal change in temperature,

Δθ of first bar = Δθ of the second bar

ΔV ÷ (2γ$V_{1}$)     =   ΔV ÷ ($V_{2}$γ)

So that;

(1 ÷ 2$V_{1}$) = (1 ÷ $V_{2}$)

2$V_{1}$ =  $V_{2}$

$V_{1}$ = $\frac{V_{2} }{2}$

Thus, original volume of the first bar is half of the original volume of the second bar.