A cube of side 5 cm made of iron and having a mass of 1500 g is heated from `25^(@)` C to `400^(@)` C. The specific heat for iron is 0.12 cal/g°C and the coefficient of volume expansion is `3.5 xx 10^(-5) //^(@) C` , the change in the internal energy of the cube is (atm pressure = `1 xx 10^(5) N//m^(2)`)

Options

A. 320 kJ
B. 282 kJ
C. 141 kJ
D. 423 kJ

Explanation

To find the change in the internal energy of the iron cube, we will follow these steps:

### Step 1: Calculate the heat absorbed (ΔQ)
The heat absorbed by the cube can be calculated using the formula:
\[ \Delta Q = m \cdot c \cdot \Delta T \]
where:
- \( m = 1500 \, \text{g} \) (mass of the cube),
- \( c = 0.12 \, \text{cal/g°C} \) (specific heat of iron),
- \( \Delta T = 400°C - 25°C = 375°C \) (temperature change).

First, we convert the specific heat from calories to joules:
\[ 1 \, \text{cal} = 4.184 \, \text{J} \]
Thus,
\[ c = 0.12 \, \text{cal/g°C} \times 4.184 \, \text{J/cal} = 0.50208 \, \text{J/g°C} \]

Now, substituting the values:
\[ \Delta Q = 1500 \, \text{g} \cdot 0.50208 \, \text{J/g°C} \cdot 375°C \]
\[ \Delta Q = 1500 \cdot 0.50208 \cdot 375 = 282,600 \, \text{J} \, \text{(or 282.6 kJ)} \]

### Step 2: Calculate the work done (W)
The work done due to volume expansion can be calculated using:
\[ W = P \cdot \Delta V \]
where:
- \( P = 1 \times 10^5 \, \text{N/m}^2 \) (atmospheric pressure),
- \( \Delta V \) is the change in volume.

To find \( \Delta V \), we use the formula:
\[ \Delta V = V_0 \cdot \beta \cdot \Delta T \]
where:
- \( V_0 \) is the initial volume,
- \( \beta = 3.5 \times 10^{-5} \, \text{°C}^{-1} \) (coefficient of volume expansion),
- \( \Delta T = 375°C \).

The initial volume \( V_0 \) of the cube is:
\[ V_0 = \text{side}^3 = (0.05 \, \text{m})^3 = 1.25 \times 10^{-3} \, \text{m}^3 \]

Now substituting the values:
\[ \Delta V = 1.25 \times 10^{-3} \, \text{m}^3 \cdot 3.5 \times 10^{-5} \, \text{°C}^{-1} \cdot 375°C \]
\[ \Delta V = 1.25 \times 10^{-3} \cdot 0.013125 = 1.64 \times 10^{-5} \, \text{m}^3 \]

Now calculating the work done:
\[ W = 1 \times 10^5 \, \text{N/m}^2 \cdot 1.64 \times 10^{-5} \, \text{m}^3 = 1.64 \, \text{J} \]

### Step 3: Calculate the change in internal energy (ΔU)
Using the first law of thermodynamics:
\[ \Delta U = \Delta Q - W \]
Substituting the values:
\[ \Delta U = 282600 \, \text{J} - 1.64 \, \text{J} \approx 282598.36 \, \text{J} \]
Converting to kJ:
\[ \Delta U \approx 282.6 \, \text{kJ} \]

### Final Answer
The change in the internal energy of the cube is approximately **282.6 kJ**.

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