Final answer:
The amount of heat energy released when 50.0 grams of water is cooled from 20.0°C to 10.0°C is 2092 Joules, calculated using the specific heat capacity of water.
Step-by-step explanation:
To find the amount of heat energy released by cooling water, you can use the specific heat capacity formula:
Q = m × c × ΔT
where:
- Q is the heat energy (in Joules),
- m is the mass of water (in grams),
- c is the specific heat capacity of water (4.184 J/g°C), and
- ΔT is the change in temperature (in °C).
Given that the mass (m) is 50.0 grams and the temperature change (ΔT) is 10.0°C (20.0°C to 10.0°C), the calculation is as follows:
Q = 50.0 g × 4.184 J/g°C × (20.0°C - 10.0°C)
Q = 50.0 g × 4.184 J/g°C × 10.0°C
Q = 2092 J
Therefore, 2092 Joules of heat energy is released when 50.0 grams of water is cooled from 20.0°C to 10.0°C.