A 0.05 kg metal sample at 200.0°C is dropped into 0.40 kg of water at 20.0°C. The final equilibrium temperature is 22.4°C. Calculate the specific heat capacity of the metal. The specific heat capacity

For the metal:
Mass, $m_m = 0.05 \text{ kg}$
Initial temperature, $T_{m,i} = 200.0 \text{ °C}$
Final equilibrium temperature, $T_f = 22.4 \text{ °C}$
Change in temperature, $\Delta T_m = T_f - T_{m,i} = 22.4 \text{ °C} - 200.0 \text{ °C} = -177.6 \text{ °C}$
Specific heat capacity, $c_m = ?$

For the water:
Mass, $m_w = 0.40 \text{ kg}$
Initial temperature, $T_{w,i} = 20.0 \text{ °C}$
Final equilibrium temperature, $T_f = 22.4 \text{ °C}$
Change in temperature, $\Delta T_w = T_f - T_{w,i} = 22.4 \text{ °C} - 20.0 \text{ °C} = 2.4 \text{ °C}$
Specific heat capacity, $c_w = 4.186 \text{ kJ/(kg·°C)} = 4186 \text{ J/(kg·°C)}$

Heat lost by metal = Heat gained by water.

The calculated value $452.54 \text{ J/(kg·°C)}$ is closest to $453 \text{ J/(kg·°C)}$.

The specific heat capacity of the metal is $\boxed{\text{453 J/(kg·°C)}}$.

: Adiabatic Free Expansion

An adiabatic process is one where no heat is exchanged with the surroundings, so $Q = 0$.
A free expansion occurs when a gas expands into a vacuum. In this process, there is no external pressure for the gas to work against.

The work done by a gas is given by $W = \int P_{ext} dV$, where $P_{ext}$ is the external pressure.
For a free expansion, the external pressure $P_{ext} = 0$.
Therefore, the work done by the gas is:

The information about the temperature decrease is a characteristic of a real gas undergoing free expansion, or a non-ideal adiabatic process. However, for an ideal gas undergoing a free expansion, the work done is defined as zero because there is no external pressure.

The work done by the gas in the process is $\boxed{\text{0 J}}$.