determine the total amount of energy required to heat 25g of ice at -18 C to steam at 125 C
To determine the total amount of energy required to heat 25g of ice at -18°C to steam at 125°C, you need to calculate the energy required for each step of the phase change.
First, you need to calculate the energy required to heat the ice from -18°C to 0°C. You can use the specific heat capacity of ice to do this calculation.
The specific heat capacity of ice is 2.09 J/g°C. To calculate the energy required to heat the ice from -18°C to 0°C, you use the following formula:
Energy = mass × specific heat capacity × temperature change
Energy = 25g × 2.09 J/g°C × (0°C - (-18°C))
Calculate the result:
Energy = 25g × 2.09 J/g°C × 18°C
Next, you need to calculate the energy required to melt the ice at 0°C. This step requires the heat of fusion of ice.
The heat of fusion of ice is 334 J/g. To calculate the energy required to melt the ice, you use the following formula:
Energy = mass × heat of fusion
Energy = 25g × 334 J/g
After the ice is melted, you need to calculate the energy required to heat the water from 0°C to 100°C. You can use the specific heat capacity of water for this calculation.
The specific heat capacity of water is 4.18 J/g°C. To calculate the energy required to heat the water, use the following formula:
Energy = mass × specific heat capacity × temperature change
Energy = 25g × 4.18 J/g°C × (100°C - 0°C)
Finally, you need to calculate the energy required to vaporize the water at 100°C. This step requires the heat of vaporization of water.
The heat of vaporization of water is 2260 J/g. To calculate the energy required to vaporize the water, you use the following formula:
Energy = mass × heat of vaporization
Energy = 25g × 2260 J/g
Now, you can add up all the energies calculated in each step to get the total energy required to heat the ice to steam.
To determine the total amount of energy required to heat 25g of ice at -18°C to steam at 125°C, we need to consider several steps:
1. Heating the ice from -18°C to 0°C.
2. Melting the ice at 0°C.
3. Heating the water from 0°C to 100°C.
4. Vaporizing the water at 100°C.
5. Heating the steam from 100°C to 125°C.
Let's break down the calculations step by step:
1. Heating the ice from -18°C to 0°C:
The specific heat capacity of ice is 2.09 J/g°C.
The temperature change is 0°C - (-18°C) = 18°C.
The energy required is: 25g × 2.09 J/g°C × 18°C = 942 J.
2. Melting the ice at 0°C:
The heat of fusion for ice is 334 J/g.
The energy required is: 25g × 334 J/g = 8350 J.
3. Heating the water from 0°C to 100°C:
The specific heat capacity of water is 4.18 J/g°C.
The temperature change is 100°C - 0°C = 100°C.
The energy required is: 25g × 4.18 J/g°C × 100°C = 10,450 J.
4. Vaporizing the water at 100°C:
The heat of vaporization for water is 2260 J/g.
The energy required is: 25g × 2260 J/g = 56,500 J.
5. Heating the steam from 100°C to 125°C:
The specific heat capacity of steam is 2.03 J/g°C.
The temperature change is 125°C - 100°C = 25°C.
The energy required is: 25g × 2.03 J/g°C × 25°C = 1275 J.
To find the total amount of energy required, sum up the values from each step:
Total energy = 942 J + 8350 J + 10,450 J + 56,500 J + 1275 J = 77,517 J.
Therefore, the total amount of energy required to heat 25g of ice at -18°C to steam at 125°C is approximately 77,517 J.
q1 = heat to increase T of ice @ -18 to zero C.
q1 = mass ice x specific heat ice x (Tfinal-Tinitial). Tfinal = 0, Tinitial = -18, 0-(-18) = +18
q2 = heat to melt ice @ zero C to liquid water @ zero C.
q2 = mass ice x heat fusion.
q3 = heat to raise temperature from zero C to 100 C.
q3 = mass water x specific heat water x (Tfinal-Tinitial).
q4 = heat to turn liquid water @ 100 C to steam @ 100 C.
q4 = mass water x heat vaporization.
q5 = heat to increase T of steam at 100 C to 125 C.
q5 = mass steam x specific heat steam x (Tfinal-Tinitial)
Total heat required = q1 + q2 + q3 + q4 + q5.