A thermally insulated, closed copper vessel contains water at `15^@C`. When the vessel is shaken vigorously for 15 minuts, the temperature rises to `17^@C`. The mass of the vessel is 100 g and that of the water is 200 g. The specific heat capacities of copper and water are `420(J)/(kg-K)` and `4200(J)/(kg-K)` resprectively. Neglect any thermal expansion.
(a) How much heat is transferred to the liquid vessel system?
(b) How much work has been doen on this system?
(c ) How much is the increase in internal energy of the system?

Explanation

To solve the problem step by step, we will break it down into three parts as per the question.

### Given Data:
- Initial temperature of water, \( T_i = 15^\circ C \)
- Final temperature of water, \( T_f = 17^\circ C \)
- Mass of the copper vessel, \( m_c = 100 \, g = 0.1 \, kg \)
- Mass of water, \( m_w = 200 \, g = 0.2 \, kg \)
- Specific heat capacity of copper, \( c_c = 420 \, \frac{J}{kg \cdot K} \)
- Specific heat capacity of water, \( c_w = 4200 \, \frac{J}{kg \cdot K} \)

### Part (a): Heat Transferred to the Liquid-Vessel System

Since the vessel is thermally insulated, there is no heat transfer to or from the surroundings. Therefore, the heat transferred to the system is zero.

\[ Q = 0 \, J \]

### Part (b): Work Done on the System

The work done on the system can be calculated using the formula for the heat gained by the copper vessel and the water.

1. Calculate the change in temperature:
\[ \Delta T = T_f - T_i = 17^\circ C - 15^\circ C = 2 \, K \]

2. Calculate the heat gained by the copper vessel:
\[ Q_c = m_c \cdot c_c \cdot \Delta T = 0.1 \, kg \cdot 420 \, \frac{J}{kg \cdot K} \cdot 2 \, K = 84 \, J \]

3. Calculate the heat gained by the water:
\[ Q_w = m_w \cdot c_w \cdot \Delta T = 0.2 \, kg \cdot 4200 \, \frac{J}{kg \cdot K} \cdot 2 \, K = 1680 \, J \]

4. Total work done on the system:
\[ W = Q_c + Q_w = 84 \, J + 1680 \, J = 1764 \, J \]

### Part (c): Increase in Internal Energy of the System

According to the first law of thermodynamics, the change in internal energy (\( \Delta U \)) is given by:

\[ \Delta U = Q - W \]

Since \( Q = 0 \) (no heat transfer), we have:

\[ \Delta U = 0 - W = -W \]

Thus, the increase in internal energy is equal to the work done:

\[ \Delta U = 1764 \, J \]

### Final Answers:
(a) Heat transferred to the liquid-vessel system: \( 0 \, J \)
(b) Work done on the system: \( 1764 \, J \)
(c) Increase in internal energy of the system: \( 1764 \, J \)