Final answer:
When 771 cal of heat is added to 5.00 g of ice at -20.0 degrees Celsius, the ice first warms to 0°C then begins to melt. All the ice will melt with heat to spare, and the final temperature of the water will be 0°C.
Step-by-step explanation:
To determine the final temperature of the ice when 771 cal of heat is added to 5.00 g of ice at -20.0 degrees Celsius, we first need to know the specific heat of ice and the amount of heat required to change the temperature of ice. The specific heat of ice is approximately 0.50 cal/g°C, and the phase change from solid to liquid for water occurs at 0°C with an enthalpy of fusion of 79.8 cal/g.
First, we'll calculate how much heat is required to raise the temperature of ice from -20.0°C to 0°C:
Heat needed = mass × specific heat × temperature change = 5.00 g × 0.50 cal/g°C × (0.0°C - (-20.0°C)) = 50 cal
Since 771 cal is greater than 50 cal, the ice will reach 0°C and start to melt. The remaining heat after warming the ice will be:
771 cal - 50 cal = 721 cal
We then check how much ice can be melted with the remaining heat:
Amount of ice melted = remaining heat / heat of fusion = 721 cal / 79.8 cal/g = 9.03 g
Since the amount of ice we have is 5.00 g and it takes 721 cal to melt 9.03 g of ice, all of 5.00 g of ice will melt before all the heat is used up. Therefore, the final state of the ice will be liquid water at 0°C, which corresponds to answer option 1).