A piece of ice of mass 2kg is added at a temperature of 2C to a container. An electric heater of 1.6V, 0.1A is used to melt the ice. The heater is used for 15 minutes. Calculate the heat supplied by t

The heat supplied by the heater is given by $Q = VIt$, where $V$ is voltage, $I$ is current, and $t$ is time.
Given: $V = 1.6\,\text{V}$, $I = 0.1\,\text{A}$, $t = 15\,\text{min} = 15 \times 60\,\text{s} = 900\,\text{s}$.

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

The ice first melts at $0^\circ\text{C}$ (assuming the specific heat capacity of ice is for ice below $0^\circ\text{C}$ and the latent heat of fusion is at $0^\circ\text{C}$). The problem states the ice is added at $2^\circ\text{C}$, which is unusual for ice. Assuming it means the ice is at $0^\circ\text{C}$ and then melts, or it's already water at $2^\circ\text{C}$. Given "piece of ice of mass 2kg is added at a temperature of $2^\circ\text{C}$", this implies the ice has already melted and is water at $2^\circ\text{C}$. However, the problem also provides "latent heat capacity of ice", which suggests melting. Let's assume the ice is at $0^\circ\text{C}$ and melts, then the resulting water warms up. If the ice is at $2^\circ\text{C}$, it's already water. Given the latent heat, it's more likely the ice is at $0^\circ\text{C}$ and melts. Let's assume the initial temperature of the ice is $0^\circ\text{C}$ for melting, and then the water formed from ice warms up from $0^\circ\text{C}$. The problem states "ice ... at a temperature of $2^\circ\text{C}$", which is contradictory. I will assume the ice is at $0^\circ\text{C}$ and then melts, and the resulting water warms up from $0^\circ\text{C}$. If the ice was truly at $2^\circ\text{C}$, it would be water, and the latent heat would not apply. Given the latent heat is provided, the most reasonable interpretation is that the ice is at $0^\circ\text{C}$ and melts.

Heat to melt ice: $Q_{\text{melt}} = m_{\text{ice}} L_{\text{f}}$
Given: $m_{\text{ice}} = 2\,\text{kg}$, $L_{\text{f}} = 336000\,\text{Jkg}^{-1}$.

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

Heat to warm the melted ice (now water) from $0^\circ\text{C}$ to final temperature $T_{\text{f}}$:
$Q_{\text{warm, ice water}} = m_{\text{ice}} c_{\text{water}} (T_{\text{f}} - 0^\circ\text{C})$
Given: $c_{\text{water}} = 4200\,\text{Jkg}^{-1}\text{K}^{-1}$.

$$Q_{\text{warm, ice water}} = (2\,\text{kg})(4200\,\text{Jkg}^{-1}\text{K}^{-1})(T_{\text{f}} - 0^\circ\text{C})$$ $$Q_{\text{warm, ice water}} = 8400 T_{\text{f}}\,\text{J}$$

Heat lost by initial water: $Q_{\text{water}} = m_{\text{water}} c_{\text{water}} (T_{\text{initial, water}} - T_{\text{f}})$
Given: $m_{\text{water}} = 40\,\text{kg}$, $T_{\text{initial, water}} = 30^\circ\text{C}$.

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

Heat lost by copper calorimeter: $Q_{\text{copper}} = m_{\text{copper}} c_{\text{copper}} (T_{\text{initial, copper}} - T_{\text{f}})$
Given: $m_{\text{copper}} = 1.5\,\text{kg}$, $c_{\text{copper}} = 400\,\text{Jkg}^{-1}\text{K}^{-1}$. The initial temperature of the calorimeter is the same as the initial water, $30^\circ\text{C}$.

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

Total heat supplied = Total heat gained.
Heat supplied by heater + Heat lost by initial water + Heat lost by copper = Heat to melt ice + Heat to warm melted ice.

$$Q_{\text{heater}} + Q_{\text{water}} + Q_{\text{copper}} = Q_{\text{melt}} + Q_{\text{warm, ice water}}$$ $$144 + 168000(30 - T_{\text{f}}) + 600(30 - T_{\text{f}}) = 672000 + 8400 T_{\text{f}}$$ $$144 + (168000 + 600)(30 - T_{\text{f}}) = 672000 + 8400 T_{\text{f}}$$ $$144 + 168600(30 - T_{\text{f}}) = 672000 + 8400 T_{\text{f}}$$ $$144 + 5058000 - 168600 T_{\text{f}} = 672000 + 8400 T_{\text{f}}$$ $$5058144 - 168600 T_{\text{f}} = 672000 + 8400 T_{\text{f}}$$ $$5058144 - 672000 = 8400 T_{\text{f}} + 168600 T_{\text{f}}$$ $$4386144 = 177000 T_{\text{f}}$$ $$T_{\text{f}} = \frac{4386144}{177000}$$ $$T_{\text{f}} \approx 24.78047^\circ\text{C}$$

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

If we strictly interpret "ice of mass 2kg is added at a temperature of $2^\circ\text{C}$", it means it's already water. In that case, there is no melting phase.
Let's re-calculate with this interpretation:
Heat gained by the $2\,\text{kg}$ of water (from melted ice) from $2^\circ\text{C}$ to final temperature $T_{\text{f}}$:
$Q_{\text{warm, 2kg water}} = m_{\text{ice}} c_{\text{water}} (T_{\text{f}} - 2^\circ\text{C})$

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

Applying conservation of energy:

$$Q_{\text{heater}} + Q_{\text{water}} + Q_{\text{copper}} = Q_{\text{warm, 2kg water}}$$ $$144 + 168600(30 - T_{\text{f}}) = 8400(T_{\text{f}} - 2)$$ $$144 + 5058000 - 168600 T_{\text{f}} = 8400 T_{\text{f}} - 16800$$ $$5058144 - 168600 T_{\text{f}} = 8400 T_{\text{f}} - 16800$$ $$5058144 + 16800 = 8400 T_{\text{f}} + 168600 T_{\text{f}}$$ $$5074944 = 177000 T_{\text{f}}$$ $$T_{\text{f}} = \frac{5074944}{177000}$$ $$T_{\text{f}} \approx 28.6719^\circ\text{C}$$

Given that the latent heat of ice is provided, the first interpretation (ice at $0^\circ\text{C}$ melts and then warms up) is more likely what the question intended, despite the confusing wording "ice ... at a temperature of $2^\circ\text{C}$". It's a common error in problem statements. I will proceed with the first interpretation.

Final Answer based on the assumption that "ice at $2^\circ\text{C}$" means ice at $0^\circ\text{C}$ that melts, and then the resulting water warms up from $0^\circ\text{C}$.

Heat to melt ice at $0^\circ\text{C}$:

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

Heat to warm melted ice (water) from $0^\circ\text{C}$ to $T_{\text{f}}$:

$$Q_{\text{warm, ice water}} = m_{\text{ice}} c_{\text{water}} (T_{\text{f}} - 0^\circ\text{C}) = (2\,\text{kg})(4200\,\text{Jkg}^{-1}\text{K}^{-1})(T_{\text{f}}) = 8400 T_{\text{f}}\,\text{J}$$

Total heat gained by ice:

$$Q_{\text{ice total}} = 672000 + 8400 T_{\text{f}}$$

Heat lost by initial water from $30^\circ\text{C}$ to $T_{\text{f}}$:

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

Heat lost by copper calorimeter from $30^\circ\text{C}$ to $T_{\text{f}}$:

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

Total heat lost by calorimeter and initial water:

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

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

$$Q_{\text{heater}} + Q_{\text{loss total}} = Q_{\text{ice total}}$$ $$144 + (5058000 - 168600 T_{\text{f}}) = 672000 + 8400 T_{\text{f}}$$ $$5058144 - 168600 T_{\text{f}} = 672000 + 8400 T_{\text{f}}$$ $$5058144 - 672000 = 8400 T_{\text{f}} + 168600 T_{\text{f}}$$ $$4386144 = 177000 T_{\text{f}}$$ $$T_{\text{f}} = \frac{4386144}{177000}$$ $$T_{\text{f}} \approx 24.78047$$

Rounding to two decimal places:

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