Heat energy of 10,000 J is supplied to a metal block of mass 600 g at atmospheric pressure. The initial temperature of the block is `20^@C`, specific heat of metal `= 300 J kg^(-1) C^(-1) ` , relative density of metal = 7.0, coefficient of volume expansion of metal `= 6 xx 10^(-5) ""^@C^(-1)` and atmospheric pressure, `P = 10^5` Pa. Also, Heat energy supplied = mass x specific heat x change in temperature
Find the change in internal energy if it is calculated as heat energy supplied minus the work done by the block on surroundings.

Options

A. 10,000.051 J
B. 1999.51 J
C. 9999.95 J
D. 10999.95 J

Explanation

To find the change in internal energy of the metal block, we will follow these steps:

### Step 1: Identify the Given Data
- Heat energy supplied (Q) = 10,000 J
- Mass of the block (m) = 600 g = 0.6 kg (conversion from grams to kilograms)
- Initial temperature (T1) = 20 °C
- Specific heat of metal (C) = 300 J/(kg·°C)
- Coefficient of volume expansion (γ) = 6 × 10^(-5) °C^(-1)
- Atmospheric pressure (P) = 10^5 Pa

### Step 2: Calculate the Final Temperature (T2)
Using the formula for heat energy supplied:
\[ Q = m \cdot C \cdot (T2 - T1) \]
Rearranging gives:
\[ T2 = \frac{Q}{m \cdot C} + T1 \]

Substituting the values:
\[ T2 = \frac{10,000}{0.6 \cdot 300} + 20 \]
\[ T2 = \frac{10,000}{180} + 20 \]
\[ T2 = 55.56 + 20 \]
\[ T2 \approx 75.56 °C \]

### Step 3: Calculate the Change in Volume (ΔV)
The change in volume due to temperature change can be calculated using:
\[ ΔV = γ \cdot V \cdot ΔT \]
Where:
- \( ΔT = T2 - T1 = 75.56 - 20 = 55.56 °C \)

First, we need to find the volume (V) of the block:
Using the density (ρ) calculated from relative density:
\[ ρ = \text{Relative Density} \times 1000 = 7 \times 1000 = 7000 \text{ kg/m}^3 \]
Now, volume (V) can be calculated as:
\[ V = \frac{m}{ρ} = \frac{0.6}{7000} \approx 8.57 \times 10^{-5} \text{ m}^3 \]

Now, substituting into the volume change formula:
\[ ΔV = 6 \times 10^{-5} \cdot (8.57 \times 10^{-5}) \cdot 55.56 \]
\[ ΔV \approx 2.857 \times 10^{-8} \text{ m}^3 \]

### Step 4: Calculate the Work Done (ΔW)
The work done by the block on the surroundings is given by:
\[ ΔW = P \cdot ΔV \]
Substituting the values:
\[ ΔW = 10^5 \cdot 2.857 \times 10^{-8} \]
\[ ΔW \approx 0.02857 \text{ J} \]

### Step 5: Calculate the Change in Internal Energy (ΔU)
Using the first law of thermodynamics:
\[ ΔU = Q - ΔW \]
Substituting the values:
\[ ΔU = 10,000 - 0.02857 \]
\[ ΔU \approx 9999.97143 \text{ J} \]

### Final Answer
The change in internal energy of the metal block is approximately:
\[ ΔU \approx 9999.97 \text{ J} \]

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