Question

How much heat is released upon converting one mole of steam (18.0 g) from 100.0 ∘C to water at 25.0 ∘C?

Answer 1

Concepts and reason

In the process of converting steam from to water at , two processes take place. Phase transition from steam to water and cooling of water from to water at .

Calculate the individual heat for each process and add the heat released to convert steam to water and heat released from the cooling of water to get the total amount of heat released for the process.

Fundamentals

The process of converting steam to water is called as the condensation process or the phase transition from the gaseous phase to liquid phase is called as the condensation process.

Write the expression to heat released for condensation process as follows:

$q=ml $

Here, is the heat released or heat absorbed, is the mass and is the latent heat of condensation.

Heat is transferred from the higher temperature to lower temperature is called as the thermal energy transfer.

The relation between amount of heat added to a system and the temperature change is as follows:

$q=mCAT $

Here, is the quantity of heat added or loss, is the mass, is the specific heat capacity, is the temperature change.

From literature, the latent heat of condensation of steam to water is .

In the value of latent heat of condensation, the negative sign represents that heat is released in the process.

Calculate the amount of heat released for phase transition of steam to water as follows:

Thus, the amount of heat released for phase transition of steam to water is .

The specific heat capacity of water is .

Calculate amount of heat released for the process of cooling of water from to as follows:

Thus, the amount of heat released for the process of cooling of water from to is .

So, the total amount of heat released in the process is .

Ans:

The total amount of heat released in the process is .