The heat of fusion of water is 335 J/g, the heat of vaporization of water is 2.26 kJ/g, the specifc heat of ice is 2.05 J/deg/g, the specific heat of steam is 2.08 J/deg/g and the specific heat of liquid water is 4.184 J/deg/g. How much heat would be needed to convert 10.73 g of ice at -16 °C to steam at 149 °C?
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To calculate the total heat required to convert the ice to steam, we need to consider the different stages of the phase change and the temperature changes involved.
First, we need to calculate the heat required to raise the temperature of the ice from -16 °C to 0 °C. We can use the specific heat capacity of ice:
Q1 = m * c_ice * ΔT1
Where:
Q1 = heat required to raise the temperature of ice to its melting point
m = mass of ice = 10.73 g
c_ice = specific heat capacity of ice = 2.05 J/deg/g
ΔT1 = change in temperature = 0 °C - (-16 °C) = 16 °C
Q1 = 10.73 g * 2.05 J/deg/g * 16 °C
Q1 = 349.472 J
Next, we need to calculate the heat required to melt the ice at its melting point (0 °C) and convert it to water at the same temperature. This is known as the heat of fusion.
Q2 = m * ΔH_fusion
Where:
Q2 = heat required for phase change (fusion)
m = mass of ice = 10.73 g
ΔH_fusion = heat of fusion of water = 335 J/g
Q2 = 10.73 g * 335 J/g
Q2 = 3592.55 J
Now, we need to calculate the heat required to raise the temperature of the water from 0 °C to 100 °C. We can use the specific heat capacity of liquid water:
Q3 = m * c_water * ΔT2
Where:
Q3 = heat required to raise the temperature of water to its boiling point
m = mass of water = mass of ice = 10.73 g
c_water = specific heat capacity of water = 4.184 J/deg/g
ΔT2 = change in temperature = 100 °C - 0 °C = 100 °C
Q3 = 10.73 g * 4.184 J/deg/g * 100 °C
Q3 = 4458.2792 J
Next, we need to calculate the heat required to convert the water at its boiling point (100 °C) to steam at the same temperature. This is known as the heat of vaporization.
Q4 = m * ΔH_vaporization
Where:
Q4 = heat required for phase change (vaporization)
m = mass of water = mass of ice = 10.73 g
ΔH_vaporization = heat of vaporization of water = 2.26 kJ/g = 2260 J/g
Q4 = 10.73 g * 2260 J/g
Q4 = 24211.8 J
Finally, we need to calculate the heat required to raise the temperature of steam from 100 °C to 149 °C. We can use the specific heat capacity of steam:
Q5 = m * c_steam * ΔT3
Where:
Q5 = heat required to raise the temperature of steam to the desired temperature
m = mass of steam = mass of water = mass of ice = 10.73 g
c_steam = specific heat capacity of steam = 2.08 J/deg/g
ΔT3 = change in temperature = 149 °C - 100 °C = 49 °C
Q5 = 10.73 g * 2.08 J/deg/g * 49 °C
Q5 = 1095.0576 J
Now, we can calculate the total heat required to convert the ice to steam:
Total heat (Q_total) = Q1 + Q2 + Q3 + Q4 + Q5
Q_total = 349.472 J + 3592.55 J + 4458.2792 J + 24211.8 J + 1095.0576 J
Q_total = 32706.1598 J
Therefore, approximately 32,706.16 J of heat would be needed to convert 10.73 g of ice at -16 °C to steam at 149 °C.