Problem Parameters and Objective

Determine the mass of water (\(m\)) in kg. Given: \(5\text{ g}\) of ice at \(-20~^\circ\text{C}\) is mixed with \(m\text{ kg}\) of water at \(30~^\circ\text{C}\). The final mixture temperature is \(0~^\circ\text{C}\). Heat exchange is only between ice and water.

• Mass of ice (\(m_{ice}\)): \(5 \text{ g} = 0.005 \text{ kg}\)
• Initial ice temperature (\(T_{ice,i}\)): \(-20~^\circ\text{C}\)
• Initial water temperature (\(T_{water,i}\)): \(30~^\circ\text{C}\)
• Final mixture temperature (\(T_f\)): \(0~^\circ\text{C}\)
• Mass of water (\(m_{water}\)): \(m \text{ kg}\)

Calorimetry Principle and Constants

Principle: Heat lost = Heat gained. Use standard physical constants:

• Specific heat of ice (\(c_{ice}\)): \(2100 \text{ J/kg}^\circ\text{C}\)
• Latent heat of fusion of ice (\(L_f\)): \(334000 \text{ J/kg}\)
• Specific heat of water (\(c_{water}\)): \(4200 \text{ J/kg}^\circ\text{C}\)

Calculating Heat Gained by Ice

Heat absorbed by ice to reach \(0~^\circ\text{C}\) and melt:

1. Heat to warm ice from \(-20~^\circ\text{C}\) to \(0~^\circ\text{C}\): \(Q_{heat\_ice} = m_{ice} \times c_{ice} \times (T_f - T_{ice,i})\) \(Q_{heat\_ice} = 0.005 \text{ kg} \times 2100 \text{ J/kg}^\circ\text{C} \times (0~^\circ\text{C} - (-20~^\circ\text{C}))\) \(Q_{heat\_ice} = 0.005 \times 2100 \times 20 = 210 \text{ J}\)
2. Heat to melt ice at \(0~^\circ\text{C}\): \(Q_{melt\_ice} = m_{ice} \times L_f\) \(Q_{melt\_ice} = 0.005 \text{ kg} \times 334000 \text{ J/kg}\) \(Q_{melt\_ice} = 1670 \text{ J}\)
3. Total heat gained (\(Q_{gain}\)): \(Q_{gain} = Q_{heat\_ice} + Q_{melt\_ice}\) \(Q_{gain} = 210 \text{ J} + 1670 \text{ J} = 1880 \text{ J}\)

Calculating Heat Lost by Water

Heat lost by water cooling from \(30~^\circ\text{C}\) to \(0~^\circ\text{C}\):

\(Q_{lost\_water} = m_{water} \times c_{water} \times (T_{water,i} - T_f)\) \(Q_{lost\_water} = m \text{ kg} \times 4200 \text{ J/kg}^\circ\text{C} \times (30~^\circ\text{C} - 0~^\circ\text{C})\) \(Q_{lost\_water} = m \times 4200 \times 30 = 126000 \times m \text{ J}\)

Solving for Water Mass (\(m\))

Equate the heat gained by the ice to the heat lost by the water:

\(Q_{gain} = Q_{lost\_water}\) \(1880 \text{ J} = 126000 \times m \text{ J}\)

Solving for \(m\):

\(m = \frac{1880}{126000}\)

The calculation yields \(m \approx 0.015 \text{ kg}\).