Question

1. How many kilojoules of energy are needed to convert 225 grams of water

at -30 C° to water at 65 C°? (3 steps-water to freezing, phase change, water
to 75 C°)

Answer 1

Approximately 167.5 kilojoules (kJ) of energy are needed to convert 225 grams of water at -30°C to water at 65°C. This can be calculated by considering three steps: raising the temperature from -30°C to 0°C, melting the ice at 0°C, and raising the temperature from 0°C to 65°C.

Step-by-step explanation:

To calculate the amount of energy needed to convert 225 grams of water at -30°C to water at 65°C, we need to consider three separate steps:

1. First, we need to calculate the energy required to raise the temperature of the water from -30°C to 0°C. The specific heat capacity of water is 4.184 J/g°C. Therefore, the energy required is: 225 g × 4.184 J/g°C × (0°C - (-30°C)) = 28,020 J.
2. Next, we need to calculate the energy required to melt the ice at 0°C. The specific heat of fusion for water is 334 J/g. So, the energy required is: 225 g × 334 J/g = 75,150 J.
3. Finally, we need to calculate the energy required to raise the temperature of the water from 0°C to 65°C. Using the specific heat capacity of water again, the energy required is: 225 g × 4.184 J/g°C × (65°C - 0°C) = 64,289 J.

To find the total energy required, we add up the energies from each step: 28,020 J + 75,150 J + 64,289 J = 167,459 J. Therefore, approximately 167.5 kilojoules (kJ) of energy are needed to convert 225 grams of water at -30°C to water at 65°C.