Calculate the heat supplied by the electric heater.

The heat supplied by the heater is given by the formula $Q = VIt$.
Given:
Voltage $V = 1.6\,\text{V}$
Current $I = 0.1\,\text{A}$
Time $t = 15\,\text{min} = 15 \times 60\,\text{s} = 900\,\text{s}$

$$Q_{\text{heater}} = (1.6\,\text{V})(0.1\,\text{A})(900\,\text{s})$$ $$Q_{\text{heater}} = 144\,\text{J}$$

The problem states "a piece of ice of mass 2kg is added at a temperature of $2^\circ\text{C}$" and also provides the "latent heat capacity of ice". Ice cannot exist at $2^\circ\text{C}$. This is a common ambiguity in physics problems. We will assume the intention is that the ice is at $0^\circ\text{C}$, melts, and then the resulting water warms up. If it were already water at $2^\circ\text{C}$, the latent heat would not be relevant.

First, calculate the heat required to melt the ice at $0^\circ\text{C}$:

$$Q_{\text{melt}} = m_{\text{ice}} L_{\text{f}}$$

Given:
Mass of ice $m_{\text{ice}} = 2\,\text{kg}$
Latent heat of fusion of ice $L_{\text{f}} = 336000\,\text{Jkg}^{-1}$

$$Q_{\text{melt}} = (2\,\text{kg})(336000\,\text{Jkg}^{-1})$$ $$Q_{\text{melt}} = 672000\,\text{J}$$

Next, calculate the heat required to warm the melted ice (now water) from $0^\circ\text{C}$ to the final temperature $T_{\text{f}}$:

$$Q_{\text{warm, ice water}} = m_{\text{ice}} c_{\text{water}} (T_{\text{f}} - 0^\circ\text{C})$$

Given:
Specific heat capacity of water $c_{\text{water}} = 4200\,\text{Jkg}^{-1}\text{K}^{-1}$

$$Q_{\text{warm, ice water}} = (2\,\text{kg})(4200\,\text{Jkg}^{-1}\text{K}^{-1})(T_{\text{f}})$$ $$Q_{\text{warm, ice water}} = 8400 T_{\text{f}}\,\text{J}$$

The total heat gained by the ice is:

$$Q_{\text{ice total}} = Q_{\text{melt}} + Q_{\text{warm, ice water}}$$ $$Q_{\text{ice total}} = 672000 + 8400 T_{\text{f}}$$

The initial temperature of the water and the calorimeter is $30^\circ\text{C}$.

Heat lost by the initial water:

$$Q_{\text{water loss}} = m_{\text{water}} c_{\text{water}} (T_{\text{initial, water}} - T_{\text{f}})$$

Given:
Mass of initial water $m_{\text{water}} = 40\,\text{kg}$
Initial temperature of water $T_{\text{initial, water}} = 30^\circ\text{C}$

$$Q_{\text{water loss}} = (40\,\text{kg})(4200\,\text{Jkg}^{-1}\text{K}^{-1})(30^\circ\text{C} - T_{\text{f}})$$ $$Q_{\text{water loss}} = 168000(30 - T_{\text{f}})\,\text{J}$$

Heat lost by the copper calorimeter:

$$Q_{\text{copper loss}} = m_{\text{copper}} c_{\text{copper}} (T_{\text{initial, copper}} - T_{\text{f}})$$

Given:
Mass of copper calorimeter $m_{\text{copper}} = 1.5\,\text{kg}$
Specific heat capacity of copper $c_{\text{copper}} = 400\,\text{Jkg}^{-1}\text{K}^{-1}$
Initial temperature of copper $T_{\text{initial, copper}} = 30^\circ\text{C}$

$$Q_{\text{copper loss}} = (1.5\,\text{kg})(400\,\text{Jkg}^{-1}\text{K}^{-1})(30^\circ\text{C} - T_{\text{f}})$$ $$Q_{\text{copper loss}} = 600(30 - T_{\text{f}})\,\text{J}$$

The total heat lost by the warmer components is:

$$Q_{\text{loss total}} = Q_{\text{water loss}} + Q_{\text{copper loss}}$$ $$Q_{\text{loss total}} = 168000(30 - T_{\text{f}}) + 600(30 - T_{\text{f}})$$ $$Q_{\text{loss total}} = (168000 + 600)(30 - T_{\text{f}})$$ $$Q_{\text{loss total}} = 168600(30 - T_{\text{f}})$$ $$Q_{\text{loss total}} = 5058000 - 168600 T_{\text{f}}$$

Assuming negligible heat loss to the surroundings, the total heat supplied by the heater plus the heat lost by the warmer components equals the total heat gained by the colder components.

$$Q_{\text{heater}} + Q_{\text{loss total}} = Q_{\text{ice total}}$$ $$144 + (5058000 - 168600 T_{\text{f}}) = 672000 + 8400 T_{\text{f}}$$ $$5058144 - 168600 T_{\text{f}} = 672000 + 8400 T_{\text{f}}$$

Rearrange the equation to solve for $T_{\text{f}}$:

$$5058144 - 672000 = 8400 T_{\text{f}} + 168600 T_{\text{f}}$$ $$4386144 = 177000 T_{\text{f}}$$ $$T_{\text{f}} = \frac{4386144}{177000}$$ $$T_{\text{f}} \approx 24.78047^\circ\text{C}$$

Rounding to two decimal places, the final temperature is:

$$T_{\text{f}} \approx 24.78^\circ\text{C}$$

The final temperature of the mixture is $\text{24.78}^\circ\text{C}$.