A metal of mass 1 kg at constant atmospheric pressure and at initial temperature `20^@C` is given a heat of 20000J. Find the following
(a) change in temperature,
(b) work done and
(c) change in internal energy.
(Given, specific heat `=400J//kg-^@C`, cofficient of cubical expansion, `gamma=9xx10^-5//^@C`, density `rho=9000kg//m^3`, atmospheric pressure `=10^5N//m^2`)

Explanation

To solve the problem step by step, we will calculate the change in temperature, work done, and change in internal energy based on the given data.

### Given Data:
- Mass of the metal (m) = 1 kg
- Initial temperature (T_initial) = 20 °C
- Heat supplied (Q) = 20,000 J
- Specific heat (c) = 400 J/(kg·°C)
- Coefficient of cubical expansion (γ) = 9 × 10^-5 /°C
- Density (ρ) = 9000 kg/m³
- Atmospheric pressure (P) = 10^5 N/m²

### (a) Change in Temperature (ΔT)

We can use the formula for heat transfer:
\[ Q = mc\Delta T \]

Rearranging the formula to find ΔT:
\[ \Delta T = \frac{Q}{mc} \]

Substituting the known values:
\[ \Delta T = \frac{20000 \, \text{J}}{1 \, \text{kg} \times 400 \, \text{J/(kg·°C)}} \]
\[ \Delta T = \frac{20000}{400} \]
\[ \Delta T = 50 \, °C \]

### (b) Work Done (W)

The work done during the expansion at constant pressure can be calculated using:
\[ W = P \Delta V \]

To find ΔV, we use the formula:
\[ \Delta V = V \cdot \gamma \cdot \Delta T \]

First, we need to find the volume (V):
\[ V = \frac{m}{\rho} = \frac{1 \, \text{kg}}{9000 \, \text{kg/m}^3} \]
\[ V = \frac{1}{9000} \, \text{m}^3 \]

Now substituting into the ΔV formula:
\[ \Delta V = \left(\frac{1}{9000} \, \text{m}^3\right) \cdot (9 \times 10^{-5}/°C) \cdot (50 \, °C) \]
\[ \Delta V = \frac{1}{9000} \cdot 4.5 \times 10^{-3} \]
\[ \Delta V = \frac{4.5 \times 10^{-3}}{9000} \]
\[ \Delta V = 5 \times 10^{-7} \, \text{m}^3 \]

Now substituting ΔV into the work done formula:
\[ W = P \Delta V = (10^5 \, \text{N/m}^2) \cdot (5 \times 10^{-7} \, \text{m}^3) \]
\[ W = 0.05 \, \text{J} \]

### (c) Change in Internal Energy (ΔU)

Using the first law of thermodynamics:
\[ \Delta U = Q - W \]

Substituting the values we have:
\[ \Delta U = 20000 \, \text{J} - 0.05 \, \text{J} \]
\[ \Delta U = 19999.95 \, \text{J} \]

### Summary of Results:
- (a) Change in Temperature (ΔT) = 50 °C
- (b) Work Done (W) = 0.05 J
- (c) Change in Internal Energy (ΔU) = 19999.95 J