Chapter 23, Problem 55A

How many calories ate given off by 1 gram of $100°C$ steam that changes phase to 1 gram of ice at $0°C$?

Solution

To Calculate: The amount of heat obtained when $1\text{gram}$ of $\text{100}°\text{C}$ stream changes its phase to $1\text{gram}$ of ice at $\text{0}°\text{C}$.

  $720\text{cal}$

Given data:

Mass of steam, $m=1\text{gram}$

Temperature of steam $=\text{100}°\text{C}$

Temperature of ice $=\text{0}°\text{C}$

Formula used:

Heat required:

  $Q=mc\Delta T$

Where,

m is mass of substance.

c is the specific heat.

  $\Delta \text{T}$is the change in temperature.

The formula to calculate required heat when there is a phase change:

  $Q=mL$

m is the mass of substance.

L is latent heat.

Calculation:

Heat lost when $100°\text{C}$of steam changes its phase to $100°\text{C}$of water is:

  $Q_1=m{L}_1$

(where,${\text{L}}_1$is latent heat of vaporization of water is equal to $540\text{cal}{\text{g}}^{-\text{1}}$)

Here

  $m=1\text{gram}$

Then,$\begin{array}{l}{Q}_1=1\times 540\\ Q_1=540\text{cal}\end{array}$

Heat lost when $100°\text{C}$water cools down to $0°C$water

  $Q_2=mc\Delta \text{T}$

(where $c$ is specific heat capacity of water is equal to $1\text{cal}/\text{g}°\text{C}$)

Here, $m=1\text{gram}$

  $Q_2=1\times 1\times \left(100-0\right)$

  $Q_2=100\text{cal}$

Heat lost when $0°\text{C}$of water changes its phase to $0°\text{C}$of ice

  $Q_3=m{L}_3$

(where, ${\text{L}}_3$is latent heat of fusion of water is equal to $80\text{cal}{\text{g}}^{-\text{1}}$and the given mass is $m=1\text{g}$)

  $\begin{array}{l}{Q}_3=1\times 80\\ =80\text{cal}\end{array}$

  $\text{Total}\text{heat}\text{lost}=Q_1+Q_2+Q_3$

  $\text{Total}\text{heat}\text{lost}=540+100+80$

  $\text{Total}\text{heat}\text{lost}=720\text{cal}$

Conclusion:

The amount of heat lost when $1\text{gram}$ of steam at $100°\text{C}$ changes its phase to 1 gram of ice at $0°\text{C}$ is $720\text{cal}$.