How much heat is necessary to change 440g of ice at -11 degrees C to water at 20 degrees C?

To find the total heat required to change 440 g of ice at -11°C to water at 20°C, we need to calculate the heat for three steps: heating the ice from -11°C to 0°C, melting the ice at 0°C to water, and heating the water from 0°C to 20°C.

1. Heating ice: ( Q_1 = m \cdot c_{ice} \cdot \Delta T = 440 , \text{g} \cdot 2.09 , \text{J/g°C} \cdot (0 - (-11)) \approx 10,065 , \text{J} )

2. Melting ice: ( Q_2 = m \cdot L_f = 440 , \text{g} \cdot 334 , \text{J/g} \approx 146,960 , \text{J} )

3. Heating water: ( Q_3 = m \cdot c_{water} \cdot \Delta T = 440 , \text{g} \cdot 4.18 , \text{J/g°C} \cdot (20 - 0) \approx 36,704 , \text{J} )

Adding these together gives ( Q_{total} \approx 10,065 + 146,960 + 36,704 \approx 193,729 , \text{J} ). Therefore, approximately 193,729 joules of heat are necessary.