How much energy must be removed from 30 L of liquid water at 0-degree celsius in order to convert it to ice?
To calculate the amount of energy that needs to be removed to convert water at 0 degree Celsius to ice, you need to use two formulas:
1. Specific Heat: Q = m * c * ΔT
2. Heat of Fusion: Q = m * Hf
Where:
Q is the amount of energy (in joules) that needs to be removed,
m is the mass of the water (in grams),
c is the specific heat capacity of water (4.186 J/g°C),
ΔT is the change in temperature (in °C),
Hf is the heat of fusion for water (334 J/g).
First, convert the given volume of water (30 L) to its mass by using the density of water, which is 1 g/mL. Therefore, 30 L of water is equal to 30 kg (30,000 g).
Next, calculate the amount of energy required to cool the water from 0°C to its freezing point, which is also 0°C (ΔT = 0°C - 0°C = 0°C).
Q1 = m * c * ΔT
Q1 = 30,000 g * 4.186 J/g°C * 0°C
Q1 = 0 J
As the change in temperature is zero, no energy is required to cool the water to its freezing point.
Finally, calculate the amount of energy needed to convert the water at its freezing point to ice:
Q2 = m * Hf
Q2 = 30,000 g * 334 J/g
Q2 = 10,020,000 J
Therefore, to convert 30 L of liquid water at 0°C to ice, you need to remove 10,020,000 Joules of energy.