# Two pieces of metal (A and B) are identical in every way, except that A has a much larger thermal...

## Question:

Two pieces of metal (A and B) are identical in every way, except that A has a much larger thermal expansion coefficient. If equal amounts of heat are added to both pieces of metal, which metal does more work on its surroundings?

1. A

2. B

3. They do the same work.

4. Unable to determine

## Thermal Expansion

Kinetic theory of matter says that temperature of a substance is the measure of its average kinetic energy of the constituent particles like atoms or molecules. Increase of kinetic energy results in increase of atomic or molecular vibration. As the atoms or molecules vibrate more vigorously, the substance expands slightly. On decreasing the temperature, the substance contracts also. The expansion and contraction of different substance are in different rates and the thermal expansion or contraction of each material is characterized by its thermal expansion coefficient. The expansion or contraction can be in length, area, or volume.

## Answer and Explanation:

**Given data**

- Two pieces of metal
*A*and*B*are identical in every way except that metal*A*has larger thermal expansion coefficient than metal piece*B*. - Equal amount of heat energy is given to both the metal pieces.

Let {eq}\Delta Q {/eq} be the amount of heat energy given to both metal pieces *A* and *B*.

Let *m* be the mass of both the metal pieces.

Let {eq}C_A = C_B = C {/eq} be the specific heat capacity for metal piece *A* and metal piece *B*.

Let {eq}\Delta T_1 = \Delta T_2 = \Delta T {/eq} be the rise in temperature of the metal pieces with given equal amount of heat energy.

Then the increase in temperature for each metal can be related to the heat energy as

{eq}\Delta Q = m C \Delta T {/eq} and {eq}\Delta Q = m C \Delta T {/eq}

This shows that thermal expansion of metal piece *A* will be larger than thermal expansion of metal piece *B*.

Let *Y* be the Young's modulus of both the metals.

Let *L* be the length of the metal pieces and *A* be the area of cross section.

The increase in length due to thermal expansion for metal *A* is {eq}\Delta L_A = L \alpha_A \Delta T {/eq}

The increase in length of metal piece *B* , {eq}\Delta L_B = L \alpha_B \Delta T {/eq}

The Young's modulus of metal piece *A* can be expressed as {eq}Y = \dfrac { S_A L } { \Delta L_A } {/eq}

Thermal stress exerted by the metal piece *A* , {eq}S_A = \dfrac { Y \Delta L_A } { L } {/eq}

Therefore work done by metal piece *A* {eq}W_A = S_A \times \Delta L_A \\ W_A = \dfrac { Y \Delta L_A } { L } \times \Delta L_A \\ W_A = Y L \alpha _A^2 \Delta T ^2 {/eq}

Similarly thermal stress exerted by metal piece *B*, {eq}S_B = \dfrac { Y \Delta L_B } { L } {/eq}

Therefore work done per area by metal piece *B*, {eq}W_B = S_B \times \Delta L_B \\ W_B = \dfrac { Y \Delta L_B } { L } \times \Delta L_B \\ W_B = Y L \alpha _B^2 \Delta T ^2 {/eq}

This shows that work done per area by metal piece *A* is greater than metal piece *B* on the surroundings. Because {eq}\alpha_A > \alpha _B {/eq}

The correct option is **option 1)**

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from General Studies Science: Help & Review

Chapter 16 / Lesson 3