Chapter 23, Problem 61A

Answer Questions 57—61 in terms of joules rather than calories.

How much energy is needed to change 1 kg of ice at $-10°C$ to steam at $120°C$?

To determine

The energy needed to change 1 kg of ice at $-{10}^{°}C$ to steam at ${120}^{°}C$ .

Answer to Problem 61A

The energy needed to change 1 kg of ice at $-{10}^{°}C$ to steam at ${120}^{°}C$ is $3.07\times {10}^{6}J$ .

Explanation of Solution

Info Mass of ice ( $m$ ) = 1 kg = 1000 g

Heat of fusion of water, $H_f$ = 334 $J{g}^{-1}$

Heat of vaporization of water, $H_v$ = 2257 $J{g}^{-1}$

Specific heat of steam, $C_{steam}$ = 2.03 $J{g}^{-1}°C^{-1}$

Specific heat of ice, $C_{ice}=2.06J{g}^{-1}°C^{-1}$

Specific heat of water, $C_{water}$ = 4.180 $J{g}^{-1}°C^{-1}$

Melting point of water = $0^{\circ }C$

Boiling point of water = $100°C$

Formula used:

  $Q=mC\Delta T$

  $Q=m{H}_f$

  $Q_{}=m{H}_v$

  $Q$= heat required to change phases or to change in temperature.

  $\Delta T$ = change in temperature

  $\Delta T=T_f-T_i$

  $T_f$ = final temperature

  $T_i =initialtemperature$

Calculation:

First step:

Heat required $\left(Q_1\right)$ to change temperature of ice from $-{10}^{°}C$ to $0^{°}C$ .

  $\Delta T=T_f-T_i$

On putting the values; we get

  $\Delta T=0^{°}C-\left(-{10}^{°}C\right)={10}^{°}C$

  $Q_1=m{C}_{ice}\Delta T$

  $Q_1=1000g\times 2.06J{g}^{-1}°C^{-1}\times 10°C=2.06\times {10}^{4}J$

  $Q_1=2.06\times {10}^{4}J$

Second step:

Heat required $\left(Q_2\right)$ to change phase from ice to liquid at $0^{\circ }C$

  $Q_2=m{H}_f=1000g\times 334J{g}^{-1}=3.34\times {10}^{5}J$

  $Q_2=3.34\times {10}^{5}J$

Third step:

Heat required $\left(Q_3\right)$ to change temperature of water from $0^{°}C$ to ${100}^{°}C$ .

  $\Delta T=T_f-T_i$

On putting the values; we get

  $\Delta T={100}^{°}C-\left(0^{°}C\right)={100}^{°}C$

  $Q_3=m{C}_{water}\Delta T$

  $Q_3=1000g\times 4.180J{g}^{-1}°C^{-1}\times 100°C=4.180\times {10}^{5}J$

  $Q_3=4.180\times {10}^{5}J$

Fourth step:

Heat required $\left(Q_4\right)$ to change phase from liquid to steam at ${100}^{\circ }C$ .

  $Q_4=m{H}_v=1000g\times 2257J{g}^{-1}=2.257\times {10}^{6}J$

  $Q_4=2.257\times {10}^{6}J$

Fifth step:

Heat required $\left(Q_5\right)$ to change temperature of water from ${100}^{°}C$ to 120 $°C$ .

  $\Delta T=T_f-T_i$

On putting the values; we get

  $\Delta T={120}^{°}C-\left({100}^{°}C\right)={20}^{°}C$

  $\begin{array}{l}{Q}_5=m{C}_{steam}\Delta T\\ Q_5=1000g\times 2.03J{g}^{-1}°C^{-1}\times 20°C=4.06\times {10}^{4}J\\ Q_5=4.06\times {10}^{4}J\end{array}$

Total heat energy$\left(Q\right)$required to needed to change 1 kg of ice at $-{10}^{°}C$to steam at ${120}^{°}C$is;

  $Q=Q_1+Q_2+Q_3+Q_4+Q_5$

On putting the values; we get

  $Q=3.07\times {10}^{6}J$

Conclusion:

Thus, the energy needed to change 1 kg of ice at $-{10}^{°}C$ to steam at ${120}^{°}C$ is $3.07\times {10}^{6}J$ .