To solve the problem of how much ice melts when 2 kg of copper is transferred from boiling water to a block of ice, we need to apply the principles of heat transfer and the concept of latent heat. Let's break this down step-by-step. ### Step 1: Understand the Heat Transfer When the hot copper is placed in contact with the ice, it will lose heat as it cools down. This heat will be absorbed by the ice, causing it to melt. The heat lost by the copper will equal the heat gained by the ice. ### Step 2: Calculate the Heat Lost by Copper The formula to calculate the heat lost (Q) by the copper is given by: \[ Q = mc\Delta T \] Where: - \( m \) = mass of the copper (2 kg) - \( c \) = specific heat capacity of copper (400 J/kg·K) - \( \Delta T \) = change in temperature Since the copper is transferred from boiling water (100°C) to a final temperature of 0°C (the temperature of the ice), the change in temperature (\( \Delta T \)) is: \[ \Delta T = T_{\text{initial}} - T_{\text{final}} = 100°C - 0°C = 100 K \] Now, substituting the values into the formula: \[ Q = 2 \, \text{kg} \times 400 \, \text{J/kg·K} \times 100 \, \text{K} \] Calculating this gives: \[ Q = 2 \times 400 \times 100 = 80000 \, \text{J} \] ### Step 3: Calculate the Mass of Ice Melted The heat gained by the ice when it melts can be calculated using the formula: \[ Q = mL \] Where: - \( m \) = mass of the ice melted (in kg) - \( L \) = latent heat of fusion of ice (3.3 x 10^5 J/kg) We know the heat gained by the ice is equal to the heat lost by the copper, so we set the two equations equal to each other: \[ 80000 \, \text{J} = m \times 3.3 \times 10^5 \, \text{J/kg} \] Now, we can solve for \( m \): \[ m = \frac{80000 \, \text{J}}{3.3 \times 10^5 \, \text{J/kg}} \] Calculating this gives: \[ m = \frac{80000}{330000} \approx 0.2424 \, \text{kg} \] ### Step 4: Convert to a Fraction To express this in terms of the options given, we can convert 0.2424 kg into a fraction: \[ m \approx \frac{80000}{330000} = \frac{8}{33} \, \text{kg} \] ### Conclusion: Correct Option Thus, the mass of ice that would melt is: **A. \( \frac{8}{33} \, \text{kg} \)** ### Step 5: Explanation of Other Options - **B. \( \frac{33}{80} \, \text{kg} \)**: This value does not match our calculated mass and is not derived from the heat transfer equations. - **C. \( \frac{80}{33} \, \text{kg} \)**: This is the reciprocal of the correct answer and does not represent a valid mass of ice melted. - **D. \( \frac{33}{8} \, \text{kg} \)**: This is also incorrect as it is significantly larger than the calculated mass and does not fit the context of the problem. ### Revision Summary - The heat lost by the copper is calculated using \( Q = mc\Delta T \). - The heat gained by the ice is calculated using \( Q = mL \). - Setting the heat lost equal to the heat gained allows us to find the mass of ice melted. - The final answer is \( \frac{8}{33} \, \text{kg} \), corresponding to option A.