How many kilojoules of energy are needed to convert 104 g of ice at -14.6 to water at 25.8°C? (The specific heat of ice at -14.6 is 2.01 J/g°C.)
To calculate the energy required to convert ice at -14.6°C to water at 25.8°C, we need to consider two main steps:
Step 1: Heating the ice from -14.6°C to 0°C
Step 2: Melting the ice at 0°C
Step 3: Heating the resulting water from 0°C to 25.8°C
Let's calculate the energy required for each step one by one:
Step 1: Heating the ice from -14.6°C to 0°C
The specific heat capacity of ice is given as 2.01 J/g°C.
The temperature change is: 0°C - (-14.6°C) = 14.6°C.
The mass is 104 g.
The energy required can be calculated using the formula:
Energy = mass × specific heat capacity × temperature change.
Energy = 104 g × 2.01 J/g°C × 14.6°C
Energy = 3057.84 J
Step 2: Melting the ice at 0°C
The energy required to melt 1 gram of ice at 0°C is called the heat of fusion.
The heat of fusion of ice is 334 J/g.
The mass is still 104 g.
The energy required can be calculated as:
Energy = mass × heat of fusion
Energy = 104 g × 334 J/g
Energy = 34736 J
Step 3: Heating the resulting water from 0°C to 25.8°C
The specific heat capacity of water is generally taken as 4.18 J/g°C.
The temperature change is: 25.8°C - 0°C = 25.8°C.
The mass is still 104 g.
The energy required can be calculated as:
Energy = mass × specific heat capacity × temperature change
Energy = 104 g × 4.18 J/g°C × 25.8°C
Energy = 10980.96 J
Now, let's sum up the energies required for each step to get the total energy:
Total Energy = Energy for step 1 + Energy for step 2 + Energy for step 3
Total Energy = 3057.84 J + 34736 J + 10980.96 J
Total Energy = 48874.8 J
Therefore, approximately 48874.8 J, or 48.9 kJ of energy, are needed to convert 104 g of ice at -14.6°C to water at 25.8°C.