How much heat energy, in kilojoules, is required to convert 36.0g of ice at 18.0C to water at 25.0C
How do you get ice at 18.0C?
To calculate the amount of heat energy needed to convert ice to water, you can use the specific heat capacity and the heat of fusion of water.
Here are the steps to follow:
Step 1: Calculate the heat energy required to raise the temperature of the ice from -18.0°C to 0°C.
- The specific heat capacity of ice is 2.09 J/g°C.
- The mass of the ice is given as 36.0 g.
- The change in temperature is 0 - (-18.0) = 18.0°C.
Q = mcΔT
Q = (36.0 g)(2.09 J/g°C)(18.0°C)
Q = 1,525.92 J
Step 2: Calculate the heat energy required to melt the ice at 0°C.
- The heat of fusion of water is 333.5 J/g.
Q = mLf
Q = (36.0 g)(333.5 J/g)
Q = 12,006 J
Step 3: Calculate the heat energy required to raise the temperature of the water from 0°C to 25.0°C.
- The specific heat capacity of water is 4.184 J/g°C.
- The mass of the water is given as 36.0 g.
- The change in temperature is 25.0°C - 0°C = 25.0°C.
Q = mcΔT
Q = (36.0 g)(4.184 J/g°C)(25.0°C)
Q = 18,000 J
Step 4: Add the heat energy values from steps 1, 2, and 3 to get the total heat energy required.
Total Heat Energy = 1,525.92 J + 12,006 J + 18,000 J
Total Heat Energy = 31,531.92 J
To convert this to kilojoules, divide the total heat energy by 1,000.
Total Heat Energy = 31,531.92 J ÷ 1,000
Total Heat Energy = 31.53 kJ
Therefore, approximately 31.53 kilojoules of heat energy is required to convert 36.0g of ice at 18.0°C to water at 25.0°C.