how much energy must be removed from 1 L of water at 25 C and turn it to ice at -10 C?

To calculate the energy that needs to be removed from the water to turn it into ice, you need to consider two factors: the energy required to lower the temperature of the water from 25°C to 0°C (the freezing point of water) and the energy required to freeze the water at 0°C.

First, let's calculate the energy required to lower the temperature from 25°C to 0°C. The specific heat capacity of water is approximately 4.18 Joules per gram per degree Celsius (J/g°C). Since we have 1 liter of water, which weighs approximately 1000 grams, we can calculate the energy:

Energy = (mass of water) × (temperature change) × (specific heat capacity)
= (1000 g) × (25°C - 0°C) × (4.18 J/g°C)

Now, let's calculate the energy required for the phase change from water at 0°C to ice at 0°C. The specific heat of fusion (or latent heat of fusion) for water is approximately 334 Joules per gram (J/g). So, we can calculate the energy:

Energy = (mass of water) × (latent heat of fusion)
= (1000 g) × (334 J/g)

Finally, let's sum up the two energies calculated above to get the total energy required:

Total energy = Energy for temperature change + Energy for phase change

So, plugging in the values:

Total energy = [(1000 g) × (25°C - 0°C) × (4.18 J/g°C)] + [(1000 g) × (334 J/g)]

Now, let's calculate the total energy:

Total energy = 1000 grams × 25 degrees Celsius × 4.18 J/g°C + 1000 grams × 334 J/g

Total energy = 104,500 Joules + 334,000 Joules

Total energy = 438,500 Joules

Therefore, approximately 438,500 Joules of energy must be removed from 1 liter of water at 25°C to turn it into ice at -10°C.