Final answer:
When 62 grams of water are cooled from 95°C to 57°C, 9795 J of heat energy are released according to the calculation using the specific heat capacity of water. However this answer does not match the provided answer options, indicating a possible typo in the question or answers.
Step-by-step explanation:
To calculate the amount of heat energy released when 62 grams of water are cooled from 95°C to 57°C, we will use the formula for heat transfer, which is Q = mcΔT, where Q is the heat energy, m is the mass of the water, c is the specific heat capacity of water (which is 4.184 J/g°C for water), and ΔT is the change in temperature. Plugging in the values gives us Q = (62 g)(4.184 J/g°C)(95°C - 57°C).
Performing the calculation, Q = (62 g)(4.184 J/g°C)(38°C) which equals Q = 62 g * 4.184 J/g°C * 38°C = 9794.912 J, which can be rounded to 9795 J. However, since this option is not listed among the answer choices, it seems there might have been a misprint or error. It's important to verify the provided answer options or to clarify the question if there seems to be a discrepancy with the answers.