To solve this problem, we need to balance the heat gained and lost in the system since heat will flow from the warmer water to the colder ice until thermal equilibrium is reached.

Known Values

1. Mass of water, $m_w = 1.5 \, \text{kg} = 1500 \, \text{g}$
2. Initial temperature of water, $T_{\text{water initial}} = 30^\circ \text{C}$
3. Temperature of ice, $T_{\text{ice}} = 0^\circ \text{C}$
4. Final temperature of the system, $T_{\text{final}} = 15^\circ \text{C}$
5. Latent heat of fusion of ice, $L_f = 80 \, \text{cal/g}$
6. Specific heat of water, c = 1 \, \text{cal/g} \, ^\circ\text{C}

Solution Strategy

1. Calculate the heat lost by the water cooling from its initial temperature to the final temperature.
2. Calculate the heat gained by the ice as it first melts, and then the resulting water warms up to the final temperature.
3. Set up an energy balance equation since heat lost by water will equal heat gained by the ice.

Step 1: Heat Lost by Water ($Q_{\text{water}}$)

Since water cools from $30^\circ \text{C}$ to $15^\circ \text{C}$: $Q_{\text{water}} = m_w \cdot c \cdot (T_{\text{initial}} - T_{\text{final}})$ $Q_{\text{water}} = 1500 \, \text{g} \cdot 1 \, \text{cal/g}^\circ\text{C} \cdot (30^\circ \text{C} - 15^\circ \text{C})$ $Q_{\text{water}} = 1500 \cdot 15 = 22500 \, \text{cal}$

Step 2: Heat Required to Melt Ice and Warm It to $15^\circ \text{C}$

Let $m_{\text{ice}}$ be the mass of ice in grams.

1. Heat required to melt the ice at 0°C: $Q_{\text{melt}} = m_{\text{ice}} \cdot L_f$ $Q_{\text{melt}} = m_{\text{ice}} \cdot 80 \, \text{cal/g}$

2. Heat required to warm the melted ice (now water) from $0^\circ \text{C}$ to $15^\circ \text{C}$: $Q_{\text{warm}} = m_{\text{ice}} \cdot c \cdot (T_{\text{final}} - T_{\text{ice}})$ $Q_{\text{warm}} = m_{\text{ice}} \cdot 1 \, \text{cal/g}^\circ\text{C} \cdot (15^\circ \text{C} - 0^\circ \text{C})$ $Q_{\text{warm}} = m_{\text{ice}} \cdot 15$

The total heat gained by the ice, $Q_{\text{ice}}$, is: $Q_{\text{ice}} = Q_{\text{melt}} + Q_{\text{warm}}$ $Q_{\text{ice}} = m_{\text{ice}} \cdot 80 + m_{\text{ice}} \cdot 15$ $Q_{\text{ice}} = m_{\text{ice}} \cdot (80 + 15) = m_{\text{ice}} \cdot 95$

Step 3: Set Up the Energy Balance

Since $Q_{\text{water}} = Q_{\text{ice}}$: $22500 = m_{\text{ice}} \cdot 95$

Solving for $m_{\text{ice}}$: $m_{\text{ice}} = \frac{22500}{95}$ $m_{\text{ice}} \approx 236.84 \, \text{g}$

Final Answer

The mass of the ice required is approximately 236.84 grams.

Would you like further details or have questions on any part of this process?

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Related Questions

1. How is latent heat different from specific heat?
2. What factors affect the amount of heat required to change the temperature of a substance?
3. How would the solution change if the initial temperature of the ice was below 0°C?
4. What is the significance of thermal equilibrium in heat transfer problems?
5. How does the specific heat of water compare to other common substances?

Tip

In heat transfer problems, carefully note the initial and final states of each substance. This helps set up an accurate energy balance.