How much heat is needed to convert 840. g of ice at -20°C to steam at 134°C?
To calculate the amount of heat needed to convert ice at -20°C to steam at 134°C, we need to consider the different phase transitions and the specific heat capacities of each substance.
First, we need to calculate the heat required to raise the temperature of ice from -20°C to its melting point at 0°C. To do this, we use the formula:
Q1 = m × c × ΔT
Where:
Q1 is the heat required (in joules),
m is the mass of the ice (840 g),
c is the specific heat capacity of ice (2.09 J/g°C), and
ΔT is the change in temperature (0°C - (-20°C) = 20°C).
Substituting the values into the formula:
Q1 = 840 g × 2.09 J/g°C × 20°C
= 35,280 J
Next, we need to calculate the heat required to melt the ice at 0°C. The specific heat of fusion (also known as the heat of fusion) for ice is 334 J/g. The formula to calculate this heat is:
Q2 = m × Hf
Where:
Q2 is the heat required (in joules),
m is the mass of the ice (840 g), and
Hf is the heat of fusion for ice (334 J/g).
Substituting the values into the formula:
Q2 = 840 g × 334 J/g
= 280,560 J
After the ice has completely melted, we need to calculate the heat required to raise the temperature of the water from 0°C to its boiling point at 100°C. The specific heat capacity for water is 4.18 J/g°C. Using the same formula as before:
Q3 = m × c × ΔT
Where:
Q3 is the heat required (in joules),
m is the mass of the water (840 g),
c is the specific heat capacity of water (4.18 J/g°C), and
ΔT is the change in temperature (100°C - 0°C = 100°C).
Substituting the values into the formula:
Q3 = 840 g × 4.18 J/g°C × 100°C
= 352,080 J
Finally, we need to account for the heat required to convert the water at 100°C to steam at 134°C. The heat of vaporization (also known as the heat of condensation) for water is 2260 J/g. The formula to calculate this heat is:
Q4 = m × Hv
Where:
Q4 is the heat required (in joules),
m is the mass of the water (840 g), and
Hv is the heat of vaporization for water (2260 J/g).
Substituting the values into the formula:
Q4 = 840 g × 2260 J/g
= 1,898,400 J
To find the total heat required, we sum up all the individual amounts of heat:
Total heat = Q1 + Q2 + Q3 + Q4
= 35,280 J + 280,560 J + 352,080 J + 1,898,400 J
= 2,566,320 J
Therefore, it would take approximately 2,566,320 joules of heat to convert 840 g of ice at -20°C to steam at 134°C.
heat ice to 0ºC
melt ice
heat water to 100ºC
boil water
heat steam to 134ºC
each step has an associated specific heat
... or heat of fusion/vaporization