Final answer:
To find the total heat released when 20.0 g of water is cooled from 30.0°C to 0.0°C and frozen to ice, we calculate the heat released during cooling and during freezing, resulting in a total of 9189.2 joules of heat released.
Step-by-step explanation:
To calculate the total amount of heat released when 20.0 g of water at 30.0°C is cooled to 0.0°C and then frozen to ice at 0.0°C, we need to take two steps. First, calculate the heat released by the water as it cools, and then calculate the heat released by the phase change from water to ice.
The heat released from cooling the water can be calculated using the formula q = m × c × ΔT, where m is the mass of the water, c is the specific heat capacity, and ΔT is the change in temperature.
The specific heat of water is 4.184 J/g °C, so:
- q = 20.0 g × 4.184 J/g°C × (30.0°C - 0.0°C)
- q = 20.0 g × 4.184 J/g°C × 30.0°C
- q = 2509.2 J
The heat released from turning water into ice (the heat of fusion) can be calculated as q = m × ΔHfus, where ΔHfus is the heat of fusion, which is 334 J/g for water:
- q = 20.0 g × 334 J/g
- q = 6680 J
Finally, we add the two amounts of heat released to find the total: 2509.2 J + 6680 J = 9189.2 J.
Therefore, 9189.2 joules of heat are released when 20.0 g of water at 30.°C is cooled to 0.0°C and frozen to ice at 0.0°C.