Final answer:
To completely melt 5.00 grams of H2O(s) at 0°C to H2O(C) at 0°C, it would require 200 kJ of heat. This includes 4.1 kJ to raise the temperature of the ice, 133.6 kJ to melt the ice, and 61.9 kJ to bring the water to a higher temperature.
Step-by-step explanation:
To completely melt 5.00 grams of H2O(s) at 0°C to H2O(C) at 0°C, you need to consider two steps: raising the temperature of the ice to its melting point and then melting the ice. To raise the temperature of the ice from -5.0°C to 0.0°C, we can use the equation Q = mcΔT, where m is the mass, c is the specific heat capacity of ice, and ΔT is the change in temperature. Using the equation, we find that it would require 4.1 kJ. To melt the ice at 0.0°C, we use the equation Q = mL, where L is the heat of fusion. Given that the heat of fusion of H2O is 79.9 cal/g, we need to convert it to Joules by multiplying by 4.184 J/cal, and then multiply it by the mass to find the total heat required for melting. Using this equation, we find that it would require 133.6 kJ. To bring the water from 0.0°C to 37°C, we can use the equation Q = mcΔT, where m is the mass, c is the specific heat capacity of water, and ΔT is the change in temperature. Plugging in the values, we find that it would require 61.9 kJ. Adding up these three values, we get a total of 200 kJ of heat required to completely melt 5.00 grams of H2O(s) at 0°C to H2O(C) at 0°C.