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At 1 atm, how much energy is required to heat 55.0g of H₂O(s) at -22.0 ^degree C to H₂O(g) at 123.0 C?

User Simon Hi
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1 Answer

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Final answer:

To heat 55.0g of H₂O(s) from -22.0 °C to H₂O(g) at 123.0 °C, you need to consider the phase changes and temperature changes. The total heat required is the sum of the heat for phase change and temperature change.

Step-by-step explanation:

To calculate the amount of energy required to heat H₂O from -22.0 °C to 123.0 °C, we need to consider the phase changes and temperature changes.


Phase Change:

First, we need to calculate the heat required to convert the ice at -22.0 °C to water at 0 °C. This can be done using the formula:

Heat = mass × heat of fusion

Heat = 55.0 g × 79.9 cal/g = 4394.5 cal


Temperature Change:

Next, we need to calculate the heat required to raise the temperature of the water from 0 °C to 123.0 °C. This can be done using the formula:

Heat = mass × specific heat × temperature change

Heat = 55.0 g × 4.184 J/g °C × (123.0 °C - 0 °C)

Heat = 28897.32 J


Total Heat Required:

The total heat required is the sum of the heat for phase change and temperature change:

Total Heat Required = 4394.5 cal + 28897.32 J

User Joel Purra
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