Final answer:
The total energy involved in changing 1 gram of liquid water at 0.0 °C to ice at -20°C is 89.8 calories, accounting for both the phase change and the subsequent temperature decrease.
Step-by-step explanation:
To calculate the energy involved in changing a sample of liquid water at 0.0 °C to ice at -20°C, we need to consider both the energy required for the phase change from liquid to solid (freezing) and the energy required to decrease the temperature of the ice to -20°C.
Firstly, the heat energy (Q) to remove from water at 0°C to turn it into ice at 0°C is calculated using the latent heat of fusion (Lf), where Q = m * Lf. Given that Lf is 79.8 cal/g for water, and assuming we have 1 gram of water, the energy required for the phase change is 79.8 cal.
Secondly, to lower the temperature of 1 gram of ice from 0°C to -20°C, we use the specific heat capacity of ice, which is 0.50 cal/g °C. The energy change for the temperature decrease (Qt) is Qt = m * c * ΔT, where m is the mass, c is the specific heat, and ΔT is the change in temperature.
Qt = 1 g * 0.50 cal/g°C * 20°C = 10 cal.
Finally, we sum the energy for the phase change and the energy for the temperature decrease to get the total energy involved.
Total energy = Energy for phase change + Energy for temperature decrease
= 79.8 cal + 10 cal = 89.8 cal