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Answer :
The amount of heat required to convert 1.70 g of water at 67.0 °C to 1.70 g of steam at 100.0 °C is 4.07 kJ.
To determine the amount of heat required to convert 1.70 g of water at 67.0 °C to steam at 100.0 °C, we need to consider the different stages of the heating process. First, we need to calculate the heat required to raise the temperature of the water from 67.0 °C to its boiling point at 100.0 °C. To do this, we can use the formula:
q = m × c × ΔT
where:
- q is the heat absorbed or released
- m is the mass of the substance
- c is the specific heat capacity of the substance
- ΔT is the change in temperature
The specific heat capacity of water is 4.18 J/g·°C.
q = 1.70 g × 4.18 J/g·°C × (100.0 °C - 67.0 °C)
q = 1.70 g × 4.18 J/g·°C × 33.0 °C
q = 225.0066 J
Now, we need to consider the heat required to convert the water at its boiling point to steam at the same temperature. This is known as the heat of vaporization. The heat of vaporization of water is 40.7 kJ/mol.
To calculate the heat required to convert 1.70 g of water to steam, we need to convert the mass of water to moles using the molar mass of water, which is approximately 18.015 g/mol.
moles of water = 1.70 g / 18.015 g/mol
moles of water = 0.0943 mol
Now we can calculate the heat required to convert the water to steam:
q = moles of water × heat of vaporization
q = 0.0943 mol × 40.7 kJ/mol
q = 3.84301 kJ
Finally, we can add the two heat values together to find the total heat required:
Total heat = heat to raise temperature + heat of vaporization
Total heat = 225.0066 J + 3.84301 kJ
Total heat = 4.06802 kJ
Therefore, the amount of heat required is approximately 4.07 kJ.
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