Explain the three modes of heat transfer and calculate the heat transferred by conduction through a wall, and find the magnetic field inside a solenoid.

Given:
Length $L = 300 \text{ cm} = 3 \text{ m}$
Area $A = 2.4 \text{ m}^2$
Initial temperature $T_1 = 30^\circ\text{C}$
Final temperature $T_2 = 90^\circ\text{C}$
Time $t = 13 \text{ s}$
Thermal conductivity $K = 200 \text{ W/mK}$

The temperature difference is $\Delta T = T_2 - T_1 = 90^\circ\text{C} - 30^\circ\text{C} = 60^\circ\text{C} = 60 \text{ K}$.
The formula for heat transfer by conduction is:

$$\frac{Q}{t} = K A \frac{\Delta T}{L}$$

Rearranging to solve for $Q$:

$$Q = K A \frac{\Delta T}{L} t$$

Substitute the given values:

$$Q = (200 \text{ W/mK}) \times (2.4 \text{ m}^2) \times \frac{60 \text{ K}}{3 \text{ m}} \times (13 \text{ s})$$ $$Q = (200 \times 2.4 \times 20 \times 13) \text{ J}$$ $$Q = (480 \times 20 \times 13) \text{ J}$$ $$Q = (9600 \times 13) \text{ J}$$ $$Q = 124800 \text{ J}$$ $$Q = 124.8 \text{ kJ}$$

The heat transferred is $\boxed{\text{124.8 kJ}}$.

Given:
Number of turns per unit length $n = 500 \text{ turns/m}$
Current $I = 2 \text{ A}$
The permeability of free space $\mu_0 = 4\pi \times 10^{-7} \text{ T}\cdot\text{m/A}$

The formula for the magnetic field inside a solenoid is:

$$B = \mu_0 n I$$

Substitute the given values:

$$B = (4\pi \times 10^{-7} \text{ T}\cdot\text{m/A}) \times (500 \text{ turns/m}) \times (2 \text{ A})$$ $$B = (4\pi \times 10^{-7} \times 1000) \text{ T}$$ $$B = (4\pi \times 10^{-4}) \text{ T}$$ $$B \approx (4 \times 3.14159 \times 10^{-4}) \text{ T}$$ $$B \approx 1.2566 \times 10^{-3} \text{ T}$$

The magnetic field inside the solenoid is $\boxed{1.26 \times 10^{-3} \text{ T}}$.

For a reversible adiabatic process, the work done $W$ is given by:

$$W = \int_{V_1}^{V_2} P dV$$

For an adiabatic process, $PV^\gamma = \text{constant} = K$. Thus, $P = K V^{-\gamma}$.
Substitute $P$ into the work integral:

$$W = \int_{V_1}^{V_2} K V^{-\gamma} dV$$ $$W = K \left[ \frac{V^{-\gamma+1}}{-\gamma+1} \right]_{V_1}^{V_2}$$ $$W = \frac{K}{1-\gamma} (V_2^{1-\gamma} - V_1^{1-\gamma})$$

Since $K = P_1 V_1^\gamma = P_2 V_2^\gamma$, we can substitute $K$:

$$W = \frac{1}{1-\gamma} (P_2 V_2^\gamma V_2^{1-\gamma} - P_1 V_1^\gamma V_1^{1-\gamma})$$ $$W = \frac{1}{1-\gamma} (P_2 V_2 - P_1 V_1)$$

For an ideal gas, $PV = nRT$. For one mole ($n=1$), $PV = RT$.
Substitute $P_1 V_1 = RT_1$ and $P_2 V_2 = RT_2$:

$$W = \frac{1}{1-\gamma} (RT_2 - RT_1)$$ $$W = \frac{R}{1-\gamma} (T_2 - T_1)$$

The magnitude of the work done is $w = |W|$, so:

$$w = \left| \frac{R}{1-\gamma} (T_2 - T_1) \right|$$

This completes the proof.

For an adiabatic process, there is no heat exchange, so $dQ = 0$.
According to the first law of thermodynamics:

$$dU = dQ - dW$$ $$dU = -dW$$

For an ideal gas, the change in internal energy is $dU = n C_v dT$, and the work done is $dW = P dV$.
So,

$$n C_v dT = -P dV \quad (1)$$

From the ideal gas law, $PV = nRT$. Differentiating this equation (assuming $n$ and $R$ are constant):

$$P dV + V dP = nR dT$$

Rearranging for $dT$:

$$dT = \frac{P dV + V dP}{nR} \quad (2)$$

Substitute equation (2) into equation (1):

$$n C_v \left( \frac{P dV + V dP}{nR} \right) = -P dV$$ $$\frac{C_v}{R} (P dV + V dP) = -P dV$$

We know that for an ideal gas, $R = C_p - C_v$. So, $\frac{C_v}{C_p - C_v} (P dV + V dP) = -P dV$.
Divide both sides by $C_v$:

$$\frac{1}{\frac{C_p}{C_v} - 1} (P dV + V dP) = -P dV$$

Let $\gamma = \frac{C_p}{C_v}$ (the adiabatic index).

$$\frac{1}{\gamma - 1} (P dV + V dP) = -P dV$$

Multiply both sides by $(\gamma - 1)$:

$$P dV + V dP = -(\gamma - 1) P dV$$

Rearrange the terms to group $P dV$:

$$V dP = -(\gamma - 1) P dV - P dV$$ $$V dP = -\gamma P dV$$

Now, separate the variables $P$ and $V$:

$$\frac{dP}{P} = -\gamma \frac{dV}{V}$$

Integrate both sides:

$$\int \frac{dP}{P} = -\gamma \int \frac{dV}{V}$$ $$\ln P = -\gamma \ln V + \ln C'$$

where $\ln C'$ is the integration constant.

$$\ln P + \gamma \ln V = \ln C'$$

Using logarithm properties, $\gamma \ln V = \ln V^\gamma$:

$$\ln P + \ln V^\gamma = \ln C'$$ $$\ln (P V^\gamma) = \ln C'$$

Exponentiate both sides:

$$P V^\gamma = C'$$

Thus, $P V^\gamma = \text{constant}$. This also implies $P_1 V_1^\gamma = P_2 V_2^\gamma$.
This completes the proof.