A thermally insulated, closed copper vessel contains water at `15^(0)C`. When the vessel is shaken vigorously for 15 minutes, the temprature rises to `17^(0)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 Jkg^(-1) K^(-1)` and `4200 J kg^(-1)K^(-1)` respectively. negalect any thermal expansion.
(a) how much heat is transferred to the liquid-vessel system ?
(b)how much work has been done on this system?
(c ) how much is the increase in internal energy if the system ?

Explanation

To solve the problem step by step, we will address each part of the question sequentially.

### 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, \( m_{copper} = 100 \, g = 0.1 \, kg \)
- Mass of water, \( m_{water} = 200 \, g = 0.2 \, kg \)
- Specific heat capacity of copper, \( c_{copper} = 420 \, J/(kg \cdot K) \)
- Specific heat capacity of water, \( c_{water} = 4200 \, J/(kg \cdot K) \)
- Change in temperature, \( \Delta T = T_f - T_i = 17 - 15 = 2 \, K \)

### Part (a): Heat Transferred to the Liquid-Vessel System
Since the system is thermally insulated, there is no heat transfer into or out of the system. Therefore, the heat transferred \( Q \) is:

\[ Q = 0 \, J \]

### Part (b): Work Done on the System
According to the first law of thermodynamics, the change in internal energy \( \Delta U \) is related to the heat transferred \( Q \) and the work done \( W \):

\[ \Delta U = Q + W \]

Since \( Q = 0 \):

\[ \Delta U = W \]

Now, we need to calculate the change in internal energy of the system, which includes both the copper vessel and the water.

The change in internal energy for the copper vessel and water can be calculated as follows:

\[ \Delta U = m_{copper} \cdot c_{copper} \cdot \Delta T + m_{water} \cdot c_{water} \cdot \Delta T \]

Substituting the values:

\[ \Delta U = (0.1 \, kg \cdot 420 \, J/(kg \cdot K) \cdot 2 \, K) + (0.2 \, kg \cdot 4200 \, J/(kg \cdot K) \cdot 2 \, K) \]

Calculating each term:

\[ \Delta U_{copper} = 0.1 \cdot 420 \cdot 2 = 84 \, J \]
\[ \Delta U_{water} = 0.2 \cdot 4200 \cdot 2 = 1680 \, J \]

Now, adding both contributions:

\[ \Delta U = 84 \, J + 1680 \, J = 1764 \, J \]

Thus, the work done on the system is:

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

### Part (c): Increase in Internal Energy of the System
From the calculations above, we have already determined that the increase in internal energy of the system is:

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

### Summary of Results:
(a) Heat transferred to the liquid-vessel system: \( Q = 0 \, J \)
(b) Work done on the system: \( W = 1764 \, J \)
(c) Increase in internal energy of the system: \( \Delta U = 1764 \, J \)