Question 100: Thermodynamics & KTG

Question

The temperature of a metal strip having coefficient of linear expansion $\alpha$ is increased from $T_1$ to $T_2$ resulting in increase of its length by $\Delta L_1$. The temperature is further increased from $T_2$ to $T_3$ such that the increase in its length is $\Delta L_2$.
Given $T_3+T_1=2 T_2$ and $T_2-T_1=\Delta T$, the value of $\Delta L_2$ is $\_\_\_\_$ .

Accepted Answer

For linear expansion, if the temperature changes from $T_a$ to $T_b$, then the increase in length is
$$ \Delta L = L_a \alpha (T_b-T_a) $$
where $L_a$ is the length at temperature $T_a$.
Now let the length at temperature $T_1$ be $L_1$.
Then for the rise from $T_1$ to $T_2$,
$$ \Delta L_1 = L_1 \alpha (T_2-T_1) $$
Given
$$ T_2-T_1=\Delta T $$
so,
$$ \Delta L_1 = L_1 \alpha \Delta T $$
Thus the length at temperature $T_2$ becomes
$$ L_2 = L_1 + \Delta L_1 = L_1(1+\alpha \Delta T) $$
Now for the rise from $T_2$ to $T_3$, the increase in length is
$$ \Delta L_2 = L_2 \alpha (T_3-T_2) $$
We are given
$$ T_3+T_1=2T_2 $$
So,
$$ T_3-T_2 = T_2-T_1 = \Delta T $$
Hence,
$$ \Delta L_2 = L_2 \alpha \Delta T $$
Substitute $L_2 = L_1(1+\alpha \Delta T)$:
$$ \Delta L_2 = L_1(1+\alpha \Delta T)\alpha \Delta T $$
But
$$ L_1 \alpha \Delta T = \Delta L_1 $$
Therefore,
$$ \Delta L_2 = \Delta L_1(1+\alpha \Delta T) $$
So the correct answer is
$$ \boxed{\Delta L_1[1+\alpha \Delta T]} $$
Hence, Option D is correct.

Answer

$\Delta L_1\left[1+2 \alpha^2(\Delta T)^2\right]$

Answer

$$ \Delta L_1\left[1+\alpha^2(\Delta T)^2\right] $$

Answer

$$ \Delta L_1[1+2 \alpha \Delta T] $$

Answer

$$ \Delta L_1[1+\alpha \Delta T] $$

Thermodynamics & KTG

Choose the correct answer: