What energy is required to heat water at #298*K# to water vapour at #373*K# under standard conditions?

Answer 1

We need #"(i) the heat capacity of water..."#

And #"(ii), we need the enthalpy of vaporization of water...."#

You have provided neither parameter in the question. And of course were asked this question in an exam, these data WOULD have been provided.

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Answer 2

The energy required to heat water from 298 K to water vapor at 373 K under standard conditions can be calculated using the formula:

[ Q = mc\Delta T ]

Where:

  • ( Q ) is the energy required (in joules).
  • ( m ) is the mass of the water (in grams).
  • ( c ) is the specific heat capacity of water (in J/g°C).
  • ( \Delta T ) is the change in temperature (in °C).

Specific heat capacity of water is approximately (4.18 , \text{J/g°C}).

To find the energy required, we need to calculate the change in temperature (( \Delta T )) and then use it in the formula along with the mass of water.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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