How many kJ of heat are required to convert 1.00 g of ice at -25°C to steam at 125°C?
s(ice) = 2.108 J/g·°C
s(liquid water) = 4.184 J/g·°C
s(steam) = 1.996 J/g·°C
heat of fusion = 6.01 kJ/mol
heat of vaporization = 40.79 kJ/mol
https://www.jiskha.com/questions/14154/how-much-heat-is-added-to-a-10-0-g-of-ice-at-20-0-degrees-celsius-to-convert-it-to-steam
To calculate the amount of heat required to convert 1.00 g of ice at -25°C to steam at 125°C, we need to consider the different stages of the process:
1. Heating the ice from -25°C to 0°C:
We can use the specific heat capacity of ice to calculate the heat required for this stage. The specific heat capacity of ice (s) is given as 2.108 J/g·°C. The temperature change is from -25°C to 0°C, which is a difference of 25°C. So, the heat required to raise the temperature of 1.00 g of ice to 0°C is:
Q1 = mass × specific heat capacity × temperature change.
Q1 = 1.00 g × 2.108 J/g·°C × 25°C.
2. Melting the ice:
Once the ice reaches 0°C, it needs to be converted to liquid water. This process requires the heat of fusion. The heat of fusion (ΔH_fus) is given as 6.01 kJ/mol. To convert it to the required heat for 1.00 g of ice, we need to calculate the number of moles (n) of ice:
n = mass / molar mass of ice.
The molar mass of ice is approximately 18.015 g/mol. So, n = 1.00 g / 18.015 g/mol. Now we can calculate the heat required to melt the ice:
Q2 = n × ΔH_fus.
3. Heating the water from 0°C to 100°C:
We need to consider the specific heat capacity of liquid water (s) for this stage, which is given as 4.184 J/g·°C. The temperature change is from 0°C to 100°C, a difference of 100°C. The heat required to raise the temperature of 1.00 g of water to 100°C is:
Q3 = mass × specific heat capacity × temperature change.
Q3 = 1.00 g × 4.184 J/g·°C × 100°C.
4. Heating the water vapor to 125°C:
Finally, we need to consider the specific heat capacity of steam (s) for this stage, which is given as 1.996 J/g·°C. The temperature change is from 100°C to 125°C, a difference of 25°C. The heat required to raise the temperature of 1.00 g of steam to 125°C is:
Q4 = mass × specific heat capacity × temperature change.
Q4 = 1.00 g × 1.996 J/g·°C × 25°C.
To get the total heat required, we add up all the heats from each stage:
Total heat = Q1 + Q2 + Q3 + Q4.
Remember to convert the units of ΔH_fus from kJ/mol to J/g by multiplying it by the molar mass of ice.
To calculate the total heat required, we need to consider three steps:
1. Heating the ice from -25°C to 0°C (melting point):
Q1 = mass x specific heat capacity x change in temperature
Q1 = 1.00 g x 2.108 J/g·°C x (0°C - (-25°C))
Q1 = 52.70 J
2. Melting the ice at 0°C to liquid water at 0°C:
Q2 = mass x heat of fusion
Q2 = 1.00 g x 6.01 kJ/mol / (18.02 g/mol) = 0.334 kJ
3. Heating the liquid water from 0°C to 125°C:
Q3 = mass x specific heat capacity x change in temperature
Q3 = 1.00 g x 4.184 J/g·°C x (125°C - 0°C)
Q3 = 520.50 J
4. Vaporizing the liquid water at 100°C to steam at 100°C:
Q4 = mass x heat of vaporization
Q4 = 1.00 g x 40.79 kJ/mol / (18.02 g/mol) = 2.266 kJ
Total heat required:
Q_total = Q1 + Q2 + Q3 + Q4
Q_total = 52.70 J + 0.334 kJ + 520.50 J + 2.266 kJ
Q_total = 3.179 kJ
Therefore, it requires 3.179 kJ of heat to convert 1.00 g of ice at -25°C to steam at 125°C.