Final answer:
The amount of heat released when 30g of water cools from 96°C to 25°C is 8922 joules, as calculated using the specific heat capacity formula.
Step-by-step explanation:
To determine how much heat is released when 30g of water at 96°C cools to 25°C, we use the specific heat capacity formula:
Q = mcΔT
where Q is the heat energy, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature.
Given that:
- m = 30g
- c = 4.18 J/g°C (specific heat of water)
- ΔT = (25°C - 96°C) = -71°C
Plugging the values into the equation:
Q = 30g * 4.18 J/g°C * (-71°C)
Q = -8922 J
Since the value of Q is negative, this indicates that 8922 joules of heat were released as the water cooled.