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Answer :
The total energy required to convert 1.70 g of ice at -12.0°C to steam at 105°C is 5585.065 J. This includes the energy needed to heat the ice to the melting point, melt it, heat the resulting water to boiling, vaporize it, and heat the steam to the given temperature.
Energy Required to Convert Ice to Steam
To calculate the energy required to convert 1.70 g of ice at
-12.0°C to steam at 105°C, we must consider several steps involving heating, melting, and vaporizing water. Each of these steps has an associated energy requirement per gram, which must be multiplied by the mass of the ice/water to get the total energy for each step.
- Heating the ice from -12.0°C to 0°C:
(2.05 J/g°C) * (1.70 g) * (12.0°C) = 41.58 J - Melting the ice at 0°C:
(333 J/g) * (1.70 g) = 565.1 J - Heating the water from 0°C to 100°C:
(4.19 J/g°C) * (1.70 g) * (100°C) = 711.3 J - Boiling the water at 100°C:
(2500 J/g) * (1.70 g) = 4250 J - Heating the steam from 100°C to 105°C:
(2.01 J/g°C) * (1.70 g) * (5°C) = 17.085 J
The total energy required is the sum of these amounts:
41.58 J + 565.1 J + 711.3 J + 4250 J + 17.085 J = 5585.065 JTherefore, the total energy required to convert 1.70 g of ice at -12.0°C to steam at 105°C is 5585.065 J.
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