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Question: A brass rod and a steel rod are both measured at \(0^{\circ}\) C. Their lengths are found to be 150 ...

A brass rod and a steel rod are both measured at 0∘0^{\circ} C. Their lengths are found to be 150 cm and 150.2 cm. At what common temperature will their lengths be equal?
(αsteel=12×10−6/∘C,αbrass=18×10−6/∘C\alpha_{steel} = 12 \times 10^{-6}/^{\circ} C, \alpha_{brass} = 18 \times 10^{-6}/^{\circ} C)
A. 111.4∘^{\circ} C
B. 167.2∘^{\circ}C
C. 222.8∘^{\circ}C
D. 238.3∘^{\circ}C

Explanation

Solution

Both the rods upon heating will expand linearly. With the use of formula for linear thermal expansion, we will get two equations with two unknowns. Substitute for one if the variables (length) to find the temperature.

Formula used:
If at an initial temperature, the length of a rod is known as l, then upon heating, the length of the rod becomes:
l′=l[1+αΔT]l' = l[1 + \alpha \Delta T]
Where ΔT\Delta T is the difference between final and initial temperatures and α\alpha is the coefficient of linear thermal expansion.

Complete answer:
We are given:
Initial temperature = 0∘^{\circ} C;
Initial length for steel, lsteell_{steel} = 150.2 cm;
Initial length for steel, lbrassl_{brass} = 150 cm;
The coefficient for linear thermal expansion of steel αsteel=12×10−6/∘C\alpha_{steel} = 12 \times 10^{-6} / ^{\circ}C;
And brass αbrass=18×10−6/∘C\alpha_{brass} = 18 \times 10^{-6} / ^{\circ}C .

Keeping these values one by one in the formula we get:
lsteel′=150.2[1+12×10−6ΔT]l'_{steel} = 150.2[1 + 12 \times 10^{-6} \Delta T]
lbrass′=150[1+18×10−6ΔT]l'_{brass} = 150[1 + 18 \times 10^{-6} \Delta T]
There are exactly two unknowns in these two equations.
Now as the initial temperature was 0∘^{\circ} C, we could write ΔT\Delta T =T - 0 = T∘^{\circ} C. And after this temperature, the lengths of both the rods become equal. So, we have:
lsteel′=lbrass′l'_{steel} = l'_{brass} .
Therefore, upon equating these two we can get:
150.2[1+12×10−6T]=150[1+18×10−6T]150.2[1 + 12 \times 10^{-6} T] = 150[1 + 18 \times 10^{-6} T]
150.2−150=(150×18−150.2×12)×10−6150.2 - 150 = (150 \times 18 - 150.2 \times 12) \times 10^{-6}
0.2×106897.6=T\dfrac{0.2 \times 10^6}{897.6} = T
T=222.816∘T = 222.816^{\circ}C
At this temperature, given brass and steel rods have equal lengths.

Therefore, the correct answer is option (C).

Note:
Unit conversions to S.I. units like from cm to m or Celsius to Kelvin are not required here as the answer is supposed to be in Celsius. Also, cm will cancel on both sides anyway so there is no requirement for any conversion of lengths to meter. In any other question requiring the same formula, one must check if the units on both sides completely match.