Final answer:
To heat 0.800 kg of water from 0°C to 30.0°C, 100.32 kJ is required. Melting 0.800 kg of 0°C ice and then raising its temperature takes a total of 367.52 kJ. This larger amount of energy needed affirms that ice absorbs more energy due to its heat of fusion.
Step-by-step explanation:
To determine the heat transfer necessary to raise the temperature of 0.800 kg of water from 0°C to 30.0°C, we utilize the specific heat capacity of water, which is approximately 4.18 kJ/kg°C. For part (a), the calculation is as follows:
Heat transfer = mass × specific heat capacity × temperature change
Heat transfer = 0.800 kg × 4.18 kJ/kg°C × (30.0°C - 0°C)
Heat transfer = 100.32 kJ
For part (b), we need to add the heat of fusion for ice, which is 334 kJ/kg, to the heat necessary to raise the water temperature after melting. This results in two separate calculations:
Heat to melt ice = mass × heat of fusion
Heat to melt ice = 0.800 kg × 334 kJ/kg
Heat to melt ice = 267.2 kJ
Heat to raise temperature (after melting) = mass × specific heat capacity × temperature change
Heat to raise temperature = 0.800 kg × 4.18 kJ/kg°C × 30.0°C
Heat to raise temperature = 100.32 kJ
Total heat transfer for ice = Heat to melt ice + Heat to raise temperature
Total heat transfer for ice = 267.2 kJ + 100.32 kJ
Total heat transfer for ice = 367.52 kJ
The total heat absorption by ice is significantly higher due to the energy required for the phase change. This demonstrates that 0°C ice is more effective at absorbing energy because it requires additional heat to change from solid to liquid before its temperature can increase.