What is the final temperature when 200 g of ice at 0°C is added to 100 g of wate

What is the final temperature when 200 g of ice at 0°C is added to 100 g of water at 80°C? (Specific heat of water = 4.2 J/g°C) (2020)

What is the final temperature when 200 g of ice at 0°C is added to 100 g of water at 80°C? (Specific heat of water = 4.2 J/g°C) (2020)

Step 1: Identify the mass of ice and water. Ice = 200 g, Water = 100 g.
Step 2: Identify the initial temperatures. Ice = 0°C, Water = 80°C.
Step 3: Understand that heat lost by water will equal heat gained by ice.
Step 4: Write the equation for heat lost by water: Q_water = mass_water * specific_heat_water * (initial_temp_water - final_temp).
Step 5: Write the equation for heat gained by ice: Q_ice = mass_ice * specific_heat_water * (final_temp - initial_temp_ice).
Step 6: Set the heat lost by water equal to the heat gained by ice: Q_water = Q_ice.
Step 7: Substitute the known values into the equation: 100 g * 4.2 J/g°C * (80°C - final_temp) = 200 g * 4.2 J/g°C * (final_temp - 0°C).
Step 8: Simplify the equation by canceling out 4.2 J/g°C from both sides.
Step 9: Solve for final_temp by rearranging the equation.
Step 10: Calculate the final temperature, which comes out to be 20°C.

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