A glass of mass 65.0 g contains 330 g of water. The glass and water are in thermal equilibrium at 25.0 ℃.Calculate the minimum mass of ice at 0.00 ℃ to be added to the glass that will reduce its temperature to 0.00 ℃.Give your answer to an appropriate number of significant figures and with units.Assume that no heat is lost to the surroundings.                specific latent heat of fusion of ice = 3.33 × 105 J kg−1                specific heat capacity of water = 4 200 J kg−1 K−1                specific heat capacity of glass = 510 J kg−1 K−1

To solve this problem, we need to calculate the amount of heat that must be removed from the glass and water to bring their temperature down to 0.00 ℃, and then determine the mass of ice required to absorb that amount of heat.

1. Calculate the heat lost by the water:

   • Mass of water, $m_w = 330 \, \text{g} = 0.330 \, \text{kg}$
   • Initial temperature of water, $T_i = 25.0 \, \text{℃}$
   • Final temperature of water, $T_f = 0.00 \, \text{℃}$
   • Specific heat capacity of water, $c_w = 4200 \, \text{J} \, \text{kg}^{-1} \, \text{K}^{-1}$

   The heat lost by the water, $Q_w$, is given by: $Q_w = m_w \cdot c_w \cdot (T_i - T_f)$ $Q_w = 0.330 \, \text{kg} \cdot 4200 \, \text{J} \, \text{kg}^{-1} \, \text{K}^{-1} \cdot (25.0 \, \text{K} - 0.00 \, \text{K})$ $Q_w = 0.330 \cdot 4200 \cdot 25.0$ $Q_w = 34650 \, \text{J}$

2. Calculate the heat lost by the glass:

   • Mass of glass, $m_g = 65.0 \, \text{g} = 0.0650 \, \text{kg}$
   • Specific heat capacity of glass, $c_g = 510 \, \text{J} \, \text{kg}^{-1} \, \text{K}^{-1}$

   The heat lost by the glass, $Q_g$, is given by: $Q_g = m_g \cdot c_g \cdot (T_i - T_f)$ $Q_g = 0.0650 \, \text{kg} \cdot 510 \, \text{J} \, \text{kg}^{-1} \, \text{K}^{-1} \cdot (25.0 \, \text{K} - 0.00 \, \text{K})$ $Q_g = 0.0650 \cdot 510 \cdot 25.0$ $Q_g = 828.75 \, \text{J}$

3. Calculate the total heat to be removed: $Q_{\text{total}} = Q_w + Q_g$ $Q_{\text{total}} = 34650 \, \text{J} + 828.75 \, \text{J}$ $Q_{\text{total}} = 35478.75 \, \text{J}$

4. Calculate the mass of ice required to absorb this heat:

   • Specific latent heat of fusion of ice, $L_f = 3.33 \times 10^5 \, \text{J} \, \text{kg}^{-1}$

   The mass of ice, $m_i$, is given by: $Q_{\text{total}} = m_i \cdot L_f$ $m_i = \frac{Q_{\text{total}}}{L_f}$ $m_i = \frac{35478.75 \, \text{J}}{3.33 \times 10^5 \, \text{J} \, \text{kg}^{-1}}$ $m_i = 0.106 \, \text{kg}$

Therefore, the minimum mass of ice required to reduce the temperature of the glass and water to 0.00 ℃ is $0.106 \, \text{kg}$ or $106 \, \text{g}$.