Chapter 23, Problem

a.

To determine

To Show: The heat required to change an $1\text{g}$ ice cube at absolute zero to $1\text{g}$ of boiling water is about . $310\text{cal}$ .

Explanation of Solution

Given:

The specific heat capacity of ice = $0.48\text{cal/g°C}$

The latent heat of fusion = $80\text{cal/g}$

The mass of ice = $1\text{g}$

The mass of water = $1\text{g}$

Formula Used:

The energy needed is the sum of energy required to convert ice into water then raise the temperature to boil water.

The heat required to raise the temperature of ice from $-273\text{°C}\text{to }0°\text{C}$

  $Q_1=mc\Delta T$

Here,

m is the mass of ice in grams

  $c_{ice}$ is the specific heat of ice in $\text{cal/g°C}$

  $\Delta T$ temperature difference

Heat requires to convert ice into the water :

  $Q_2=m{L}_f$

  $L_f$ is the latent heat of fusion.

The heat required to change the temperature of water to boiling water:

  $Q_3=mc\Delta T$

m is the mass of ice in grams

  $c_{water}$ is the specific heat of the water in $\text{cal/g°C}$

  $\Delta T$ temperature difference

Total heat requires:

  $Q_{total}=Q_1+Q_2+Q_3$

Calculation:

As per the given problem

  $\begin{array}{l}m\text{=1}\text{g}\\ c_{ice}\text{=}0.48\text{cal/g°c}\\ c_{water}\text{=1}\text{cal/g°c}\\ L_f\text{=}\text{80}\text{cal/g}\end{array}$

  $\begin{array}{l}{Q}_1=1\times 0.48\times (273)=131.04\text{cal}\\ Q_2=1\times 80=80\text{cal}\\ Q_3=1\times 1\times 100=100\text{cal}\\ Q_{total}=131+80+100=311.04\text{cal}\end{array}$

Conclusion:

The total heat required is 311.04 $\text{cal}$ that is round off to 310 $\text{cal}$ which has to be proved.

b.

To determine

To Show: The energy required to convert $100\text{°C}$ water to water vapor is much more than part a.

Answer

The energy require is 540 cal.

Explanation of Solution

Given:

The latent heat of vaporization = $540\text{cal/g}$

The mass of water = $1\text{g}$

Formula Used:

Heat requires to convert water into vapor :

  $Q_4=m{L}_{vap}$

  $L_{vap}$ is the latent heat of fusion.

Calculation:

As per the given problem

  $\begin{array}{l}m=1\text{g}\\ L_{vap}=540\text{cal/g}\end{array}$

The energy requires to convert the $100\text{°C}$ water to $100\text{°C}$ vapors:

  $\begin{array}{l}{Q}_4=1\times 540=540\text{cal}\\ Q_4>Q_{total}\end{array}$

Conclusion:

The energy required is much more than the part a.