Final answer:
The water absorbed 2509 joules of heat energy when the temperature increased from 25 to 37°C, calculated using the specific heat formula Q = mc∆T with the provided values.
Step-by-step explanation:
To calculate the heat energy absorbed by the water as it is heated, we use the equation Q = mc∆T, where Q is the heat absorbed, m is the mass of the water, c is the specific heat of water, and ∆T is the change in temperature. Given the mass of the water (50 g), specific heat of water (4.18 J/g°C), the initial temperature (25°C), and the final temperature (37°C), we can find the temperature change (∆T) to be 12°C (37°C - 25°C).
Using the values provided:
Q = (50 g) * (4.18 J/g°C) * (12°C)
Q = 2509 J
Therefore, the water absorbed 2509 joules of heat energy as the temperature increased from 25 to 37°C.