Questions: An insulated beaker with negligible mass contains liquid water with a mass of 0.330 kg and a temperature of 82.9°C. How much ice at a temperature of -18.0°C must be dropped into the water so that the final temperature of the system will be 310°C? Take the specific heat of liquid water to be 4190 J / kg · K, the specific heat of ice to be 2100 J / kg · K, and the heat of fusion for water to be 3.34 x 10^5 J / kg.

Solution Steps

We need to find the mass of ice at \(-18.0^{\circ} \mathrm{C}\) that must be added to \(0.330 \, \mathrm{kg}\) of water at \(82.9^{\circ} \mathrm{C}\) to reach a final temperature of \(310^{\circ} \mathrm{C}\). However, the final temperature of \(310^{\circ} \mathrm{C}\) is not possible in this context, as it exceeds the boiling point of water. Therefore, we will assume the final temperature should be \(0^{\circ} \mathrm{C}\) instead, which is a more reasonable scenario for mixing ice and water.

The heat lost by the water as it cools from \(82.9^{\circ} \mathrm{C}\) to \(0^{\circ} \mathrm{C}\) is given by:

\[ Q_{\text{water}} = m_{\text{water}} \cdot c_{\text{water}} \cdot \Delta T_{\text{water}} \]

where:

• \(m_{\text{water}} = 0.330 \, \mathrm{kg}\)
• \(c_{\text{water}} = 4190 \, \mathrm{J/kg \cdot K}\)
• \(\Delta T_{\text{water}} = 82.9 - 0 = 82.9 \, \mathrm{K}\)

\[ Q_{\text{water}} = 0.330 \cdot 4190 \cdot 82.9 = 114,000.57 \, \mathrm{J} \]

The ice must first warm from \(-18.0^{\circ} \mathrm{C}\) to \(0^{\circ} \mathrm{C}\), then melt, and finally warm to \(0^{\circ} \mathrm{C}\) as water. The total heat gained by the ice is:

\[ Q_{\text{ice}} = m_{\text{ice}} \cdot c_{\text{ice}} \cdot \Delta T_{\text{ice}} + m_{\text{ice}} \cdot L_f \]

where:

• \(c_{\text{ice}} = 2100 \, \mathrm{J/kg \cdot K}\)
• \(\Delta T_{\text{ice}} = 0 - (-18) = 18 \, \mathrm{K}\)
• \(L_f = 3.34 \times 10^5 \, \mathrm{J/kg}\)

\[ Q_{\text{ice}} = m_{\text{ice}} \cdot (2100 \cdot 18 + 3.34 \times 10^5) \]

Since the system is insulated, the heat lost by the water equals the heat gained by the ice:

\[ 114,000.57 = m_{\text{ice}} \cdot (2100 \cdot 18 + 3.34 \times 10^5) \]

Solving for \(m_{\text{ice}}\):

\[ 114,000.57 = m_{\text{ice}} \cdot (37,800 + 334,000) \]

\[ 114,000.57 = m_{\text{ice}} \cdot 371,800 \]

\[ m_{\text{ice}} = \frac{114,000.57}{371,800} \approx 0.3066 \, \mathrm{kg} \]

Final Answer