Question

How much heat is released upon converting one mole of steam (18.0

Answer 1

The amount of heat released in conversion of the $1\text{ mole}$ steam into water is $\boxed{4.65\times10^4\text{ J}}$.

Explanation:

Given:

The amount of water present in $1\text{ mole}$ is $18\text{ gm}$.

The final temperature of the water is $25^\circ\text{ C}$.

Concept:

The amount of heat released in the conversion of the steam into water at $25^\circ\text{ C}$ involves the conversion of the steam at $100^\circ\text{ C}$ into the water at $100^\circ\text{ C}$ and then the water at $100^\circ\text{ C}$ releases the energy to come to $25^\circ\text{ C}$.

The energy released in the conversion of the steam into the water is called as the latent heat of vaporization.

The total energy released in the conversion of the steam into water is given by:

$\boxed{Q_{\text{total}}=m.L_v+mC\Delta T}$

Here, $Q_{\text{total}}$ is the total amount of heat, $m$ is the mass of water, $L_v$ is latent heat of vaporization of water, $C$ is the specific heat capacity of water and $\Delta T$ is the change in temperature.

The latent heat of vaporization of water is $2268\text{ J/g}$ and the specific heat capacity of water is $4.2\text{ J/g.K}$.

Substitute the values on above expression.

$\begin{aligned}Q_{\text{total}}&=18(2268)+18(4.2)(100-25)\\&=4.08\times10^4+5.67\times10^3\\&\approx4.65\times10^4\text{ J}\end{aligned}$

Thus, The amount of heat released in conversion of the $1\text{ mole}$ steam into water is $\boxed{4.65\times10^4\text{ J}}$.

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Answer Details:

Grade: Senior School

Subject: Physics

Chapter: Heat Transfer

Keywords:

heat, released, water, steam, latent heat, specific heat capacity, vaporization, one mole, conversion, 25 degree, 100 degree.

Answer 2

4.65 × 10⁴ Joules of heat is released upon converting one mole of steam to water.

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Further explanation

Specific Heat Capacity is the amount of energy needed to raise temperature of 1 kg body for 1°C.

$\large {\boxed{Q = m \times c \times \Delta t} }$

Q = Energy ( Joule )

m = Mass ( kg )

c = Specific Heat Capacity ( J / kg°C )

Δt = Change In Temperature ( °C )

Let us now tackle the problem!

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Given:

initial temperature of steam = t = 100.0°C

specific heat capacity of water = c = 4.186 J/gK

mass of steam = m = 18.0 gram

final temperature of water = t' = 25.0°C

specific latent heat of vaporization of water = Lv = 2268 J/g

Asked:

heat released = Q = ?

Solution:

$\boxed {\large {\texttt{steam 100}^oC \overset{[Q_1]}{\rightarrow} \texttt{water 100}^oC \overset{[Q_2]}{\rightarrow} \texttt{water 25}^oC}}$

$Q = Q_1 + Q_2$

$Q = mL_v + mc\Delta t$

$Q = 18.0(2268) + 18.0(4.186)(100-25)$

$Q = 40824 + 5651.1$

$Q = 46475.1 \texttt{ Joules}$

$Q \approx 4.65 \times 10^4 \texttt{ Joules}$

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Answer details

Grade: College

Subject: Physics

Chapter: Thermal Physics

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Keywords: Heat , Temperature , Block , Aluminium , Ice , Cold , Water