calculate the amount of energy (kJ) lost when 50 mL of water at 0 Celcius freezes at 0 Celcius
To calculate the amount of energy lost when water freezes, we need to use the equation:
\(Q = mcΔT\)
where:
- Q is the amount of energy (in joules) lost or gained
- m is the mass of the substance (in grams)
- c is the specific heat capacity of the substance (in J/g°C)
- ΔT is the change in temperature (in °C)
To convert the mass of water from mL to grams, we need to use the density of water, which is 1 g/mL. Assuming the density is constant, we can convert the volume to grams:
\(mass = volume \times density\)
\(mass = 50 mL \times 1 g/mL\)
\(mass = 50 g\)
Next, let's calculate the change in temperature, ΔT. Since the water is freezing at its freezing point, the change in temperature is 0°C:
\(\Delta T = 0°C - 0°C = 0°C\)
Now, we need to find the specific heat capacity (c) of water. The specific heat capacity of water is approximately 4.18 J/g°C.
Finally, we can substitute the values into the equation to calculate the energy lost (Q):
\(Q = mcΔT\)
\(Q = 50g \times 4.18 J/g°C \times 0°C\)
\(Q = 0 J\)
Therefore, no energy is lost or gained when 50 mL of water at 0°C freezes at 0°C. This is because the energy lost during the freezing process is equal to the energy gained to cool the water to its freezing point.