Final answer:
To find the heat absorbed when 198 mg of H2O melts at 0 °C, the heat of fusion for water, which is 79.9 cal/g, is used. For this mass of water, the total heat absorbed during melting is calculated as 15.802 calories.
Step-by-step explanation:
The student's question concerns the amount of heat absorbed when a certain mass of H2O (water) melts at 0 °C. To answer this question, we need to consider the heat of fusion for water, that is, the amount of heat required to convert water from solid to liquid state without changing its temperature.
The heat of fusion for water is 79.9 cal/g, or equivalently, 6.01 kJ/mol. Since the student's question involves a mass of 198 mg (0.198 g) of water, we can calculate the heat absorbed using the specific heat of fusion value.
Here is the calculation:
- First, convert the mass from grams to moles by dividing by the molar mass of H2O, which is approximately 18 g/mol.
- Then, multiply the result by the molar heat of fusion to find the total heat absorbed during the melting process.
However, as the mass given in the question (198 mg) is small, and the molar heat of fusion is often given in larger units (known as kJ/mol), it may be more practical to use the specific heat of fusion (cal/g) directly for such small masses.
Heat absorbed = mass (g) × heat of fusion (cal/g)
Heat absorbed = 0.198 g × 79.9 cal/g
Heat absorbed = 15.802 cal
Therefore, when 198 mg of H2O melts at 0 °C, 15.802 calories of heat are absorbed.