The internal energy of a solid also increases when heat is transferred to it from its surroundings. A 5 kg solid bar is heated at atmospheric pressure. Its temperature increases from `20^@C` to `70^@C`. The linear expansion coefficient of solid bar is `1xx10^(-3)//^(@)C`. The density of solid bar is `50 kg//m^3`. The specific heat capacity of solid bar is `200 J//kg C^@`. The atmospheric pressure is `1xx10 ^5N//m^2`.
The work done by the solid bar due to thermal expansion, under atmospheric pressure is

Options

A. 500 J
B. 1000 J
C. 1500 J
D. 2000 J

Explanation

To solve the problem of calculating the work done by the solid bar due to thermal expansion under atmospheric pressure, we can follow these steps:

### Step 1: Identify the given data
- Mass of the solid bar, \( m = 5 \, \text{kg} \)
- Initial temperature, \( T_1 = 20 \, ^\circ C \)
- Final temperature, \( T_2 = 70 \, ^\circ C \)
- Linear expansion coefficient, \( \alpha = 1 \times 10^{-3} \, ^\circ C^{-1} \)
- Density of the solid bar, \( \rho = 50 \, \text{kg/m}^3 \)
- Specific heat capacity, \( c = 200 \, \text{J/(kg} \cdot ^\circ C) \)
- Atmospheric pressure, \( P = 1 \times 10^5 \, \text{N/m}^2 \)

### Step 2: Calculate the change in temperature
\[ \Delta T = T_2 - T_1 = 70 \, ^\circ C - 20 \, ^\circ C = 50 \, ^\circ C \]

### Step 3: Calculate the initial volume of the solid bar
Using the formula for volume based on mass and density:
\[ V = \frac{m}{\rho} = \frac{5 \, \text{kg}}{50 \, \text{kg/m}^3} = 0.1 \, \text{m}^3 \]

### Step 4: Calculate the change in volume due to thermal expansion
The change in volume \( \Delta V \) can be calculated using the formula:
\[ \Delta V = V \cdot \gamma \cdot \Delta T \]
where \( \gamma = 3 \alpha \) (since \( \gamma \) is the cubical expansion coefficient).

Calculating \( \gamma \):
\[ \gamma = 3 \cdot \alpha = 3 \cdot (1 \times 10^{-3}) = 3 \times 10^{-3} \, ^\circ C^{-1} \]

Now substituting the values:
\[ \Delta V = V \cdot (3 \alpha) \cdot \Delta T = 0.1 \, \text{m}^3 \cdot (3 \times 10^{-3} \, ^\circ C^{-1}) \cdot (50 \, ^\circ C) \]
\[ \Delta V = 0.1 \cdot 3 \cdot 10^{-3} \cdot 50 = 0.1 \cdot 0.15 = 0.015 \, \text{m}^3 \]

### Step 5: Calculate the work done by the solid bar
The work done \( W \) due to thermal expansion at constant pressure is given by:
\[ W = P \cdot \Delta V \]
Substituting the values:
\[ W = (1 \times 10^5 \, \text{N/m}^2) \cdot (0.015 \, \text{m}^3) = 1500 \, \text{J} \]

### Final Answer
The work done by the solid bar due to thermal expansion under atmospheric pressure is \( 1500 \, \text{J} \).

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