Chapter 14, Problem 23A

Interpretation Introduction

Interpretation:

The total quantity of heat evolved is to be calculated.

Concept introduction:

Heat $\text{(q)}$ is thermal energy which is either absorbed or released between hotter and cooler object that are in contact. The heat transfer results in a temperature change of the system. This temperature change is expressed by the specific heat capacity ${\text{(C}}_{\text{S}}\text{)}$ which is defined as; the amount of energy is required to increase the temperature of $\text{1g}$ or $\text{1 mol}$ of substance by ${\text{1}}^{\text{o}}\text{C}$ or $\text{1 K}$ . It is written mathematically as:

  $Q=m\times s\times \Delta t$

To calculate the energy released to change the phase, mathematical expression is,

  $Q=n\times \Delta H_F(\text{or }\Delta H_V)$

Answer to Problem 23A

The specific heat capacity of an unknown substance is

Explanation of Solution

Given Information:

• $m\text{ = 10}\text{.0 g}$of steam Cs of ice $\text{= 2}\text{.06 }\frac{\text{J}}{\text{g}{\text{.}}^{\text{o}}\text{C}}$
• Cs of liquid water $\text{= 4}\text{.18 }\frac{\text{J}}{\text{g}{\text{.}}^{\text{o}}\text{C}}$
• Cs of steam = $\text{= 2}\text{.03 }\frac{\text{J}}{\text{g}{\text{.}}^{\text{o}}\text{C}}$
• $\text{Δ}T{\text{ = steam at 200}}^{\text{o}}{\text{C to ice at -50}}^{\text{o}}\text{C}$
• $\text{Δ}{H}_{\text{F}}\text{ = 6}\text{.02}\frac{\text{kJ}}{\text{mol}}$
• $\text{Δ}{H}_{\text{V}}\text{ = 40}\text{.6}\frac{\text{kJ}}{\text{mol}}$

The formula for heat (q) is,

  $Q\text{ = }m\times Cs\times \text{ Δ}t$

Where,

• q is the heat transferred in the system.
• m is the mass of the substance in grams.
• Cs is the specific heat capacity.
• $\text{ΔT}$ is the change in temperature during the heat transfer

The change in temperature ( $\text{ΔT}$ ) calculated by the following formula,

  $\Delta T=T_{\text{final}}-T_{initial}$

The formula for heat while phase change is,

  $Q\text{ = }n\text{ x Δ}{H}_{\text{F}}\text{(or Δ}{H}_{\text{V}}\text{)}$

Where,

• $\text{Δ}{H}_{\text{f}}\text{=}$ Heat of fusion
• $\text{Δ}{H}_{\text{V}}\text{=}$ Heat of vaporization

The heat released to change the temperature of the steam from ${\text{200}}^{\text{o}}\text{C}$ to ${\text{0}}^{\text{o}}\text{C}$ ${\text{(Q}}_{\text{1}}\text{)}$ ,

  $\text{ Q= 10}\text{.0 g x 2}\text{.03}\frac{\text{J}}{{\text{g }}^{\text{o}}\text{C}}{\text{x (200}}^{\text{o}}{\text{C - 0}}^{\text{o}}\text{C)}$

  $\text{ = 4060 J}$

  $\text{ = 4}\text{.06 kJ}$

First, the moles of $\text{10}\text{.0 g}$ steam are,

  $\text{Moles = }\frac{\text{mass}}{\text{molar mass}}$

  $\text{= }\frac{\text{10}\text{.0 g}}{\text{18}\text{.01}\frac{\text{g}}{\text{mol}}}$

  $\text{= 0}\text{.555 mol}$

During the phase change, the change in temperature is zero. The heat required to change the phase from steam to liquid ${\text{(Q}}_{\text{2}}\text{)}$ ,

  $Q_2=n\times \Delta H_V$

  $Q_2\text{ = 0}\text{.555 x 40}\text{.6}\frac{\text{kJ}}{\text{mol}}$

  $\text{ = 22}\text{.533 kJ}$

Now, the heat released to change the temperature of the water from ${\text{0}}^{\text{o}}\text{C}$ to ${\text{-50}}^{\text{o}}{\text{C (Q}}_{\text{3}}\text{)}$ ,

  $Q_3=m\times Cs\times \Delta t$

  $\text{ = 10}\text{.0g x 4}\text{.18}\frac{\text{J}}{{\text{g}}^{\text{o}}\text{C}}{\text{ x (0}}^{\text{o}}{\text{C - (-50}}^{\text{o}}\text{C))}$

  $\text{= 2090 J}$

  $\text{= 2}\text{.09 kJ}$

The heat required to change the phase change from water to ice ${\text{(Q}}_{\text{4}}\text{)}$ ,

  $Q_4=n\times \Delta H_V$

  $\text{= 0}\text{.555 mol x 6}\text{.02}\frac{\text{kJ}}{\text{mol}}$

  $\text{= 3}\text{.3411 kJ}$

Hence, the total quantity of heat is,

  $Q=Q_1+Q_2+Q_3+Q_4$

  $\text{ = 4}\text{.06 + 22}\text{.533 + 2}\text{.09 + 3}\text{.3411}$

  $\text{= 32}\text{.0241 kJ}$

Here, heat is evolved during the process. Hence total heat evolved is $\text{-32}\text{.0 kJ}$ .

Conclusion

The sign of heat $\text{(q)}$ either be positive or negative depending upon the specific heat capacity is either added or removed from a unit mass of a substance to change the temperature by ${\text{1}}^{\text{o}}\text{C}$ or $\text{1 K}$ .