Chapter 4, Problem 4.76AP

(a)

Interpretation Introduction

Interpretation:

The amount of heat energy needed to heat $150\text{g}$ of water has to be given.

Concept introduction:

The specific heat is the amount of heat energy needed to raise the temperature of $\text{1g}$ of a substance by ${\text{1}}^{\text{o}}\text{C}$.

  $\text{Specific}\text{heat=}\frac{\text{Heat}}{\text{mass×ΔT}}$

Answer to Problem 4.76AP

The amount of energy needed to heat water is $5250\text{cal}\text{.}$

The heat needed to vaporize the water is $\text{81000}\text{calories}\text{.}$

Explanation of Solution

The amount of energy needed to heat water:

Given,

The mass of water is $150\text{g}\text{.}$

The specific heat of water is $\text{1}\text{.00}\text{cal/g}{}^{\text{o}}\text{C}\text{.}$

The temperature at initial $\left({\text{T}}_{\text{1}}\right)$ is ${35}^{\text{o}}\text{C}\text{.}$

The temperature at final $\left({\text{T}}_2\right)$ is ${100}^{\text{o}}\text{C}\text{.}$

The heat of vaporization of water is $540\text{cal/g}\text{.}$

The change in temperature $\left(\text{ΔT}\right)$ can be calculated as,

  ${\text{ΔT=T}}_{\text{2}}{\text{-T}}_{\text{1}}$

  ${\text{ΔT=100}}^{\text{o}}\text{C}{\text{-35}}^{\text{o}}\text{C}$

  $\text{ΔT=}{\text{65}}^{\text{o}}\text{C}$

The amount of energy needed to heat water can be given as,

  $\text{Specific}\text{heat=}\frac{\text{Heat}}{\text{mass×ΔT}}$

  $\text{Heat}\text{=}\text{mass}\text{×ΔT}\text{×}\text{}\text{Specific}\text{heat}$

  $\text{Heat}\text{=}150{\text{g×35}}^{\text{o}}\text{C}\text{×}\text{}\text{1}\text{.00}{\text{cal/g}}^{\text{o}}\text{C}$

  $\text{Heat}\text{=}5250\text{cal}$

The amount of energy needed to heat water is $5250\text{cal}\text{.}$

The heat needed to vaporize:

The heat needed to vaporize the water can be calculated as,

  $\text{mass}\text{of}\text{substance}\left(\text{g}\right)\text{×}\frac{\text{heat}\text{of}\text{vaporization}\left(\text{cal}\right)}{\text{1g}}$

  $150\text{g×}\frac{540\text{cal}}{\text{1g}}\text{=81000}\text{calories}$

The heat needed to vaporize the water is $\text{81000}\text{calories}\text{.}$

(b)

Interpretation Introduction

Interpretation:

The amount of heat energy released when $\text{42}\text{g}$ of water is condensed and the heat released when the water is cooled has to be given.

Concept introduction:

Refer to part (a).

Answer to Problem 4.76AP

The amount of energy released when water is condensed is $\text{-}\text{4}\text{.200}\text{kcal}\text{.}$

The amount of heat energy released when $\text{42}\text{g}$ of water is frozen is $\text{3}\text{.347}\text{kcal}\text{.}$

Explanation of Solution

The amount of heat energy released when $\text{42}\text{g}$ of water is condensed:

Given,

The mass of water is $42\text{g}\text{.}$

The specific heat of water is $\text{1}\text{.00}\text{cal/g}{}^{\text{o}}\text{C}\text{.}$

The temperature at initial $\left({\text{T}}_{\text{1}}\right)$ is ${100}^{\text{o}}\text{C}\text{.}$

The temperature at final $\left({\text{T}}_2\right)$ is ${\text{0}}^{\text{o}}\text{C}\text{.}$

The heat of fusion of water is $\text{79}\text{.7}\text{cal/g}\text{.}$

The change in temperature $\left(\text{ΔT}\right)$ can be calculated as,

  ${\text{ΔT=T}}_{\text{2}}{\text{-T}}_{\text{1}}$

  $\text{ΔT=}{\text{-100}}^{\text{o}}\text{C}$

The amount of energy released when water is condensed can be given as,

  $\text{Specific}\text{heat=}\frac{\text{Heat}}{\text{mass×ΔT}}$

  $\text{Heat}\text{=}\text{mass}\text{×ΔT}\text{×}\text{}\text{Specific}\text{heat}$

  $\text{Heat}\text{=}\text{42}\text{g×-}{100}^{\text{o}}\text{C}\text{×}\text{}\text{1}\text{.00}{\text{cal/g}}^{\text{o}}\text{C}$

  $\text{Heat}\text{=}-4200\text{cal}$

The energy in $\text{cal}$ can be converted to $\text{kcal}$ as,

  $\text{1}\text{kcal=1,000}\text{cal}$

  $\text{-4200}\cancel{\text{cal}}\text{×}\frac{\text{1}\text{kcal}}{\text{1000}\cancel{\text{cal}}}\text{=}\text{-4}\text{.200}\text{kcal}$

The amount of energy released when water is condensed is $\text{-}\text{4}\text{.200}\text{kcal}\text{.}$

The amount of energy released when water if frozen to ice can be calculated as,

The amount of energy needed to froze can be given as,

  $\text{mass}\text{of}\text{substance}\left(\text{g}\right)\text{×}\frac{\text{heat}\text{of}\text{fusion}\left(\text{cal}\right)}{\text{1g}}$

  $42\text{g×}\frac{\text{79}\text{.7}\text{cal}}{\text{1g}}\text{=3347}\text{calories}$

The amount of heat energy released when $\text{42}\text{g}$ of water is frozen is $3347\text{cal}\text{.}$

The energy in $\text{cal}$ can be converted to $\text{kcal}$ as,

  $\text{1}\text{kcal=1,000}\text{cal}$

  $\text{3347}\cancel{\text{cal}}\text{×}\frac{\text{1}\text{kcal}}{\text{1000}\cancel{\text{cal}}}\text{=3}\text{.347}\text{kcal}$

The amount of heat energy released when $\text{42}\text{g}$ of water is frozen is $\text{3}\text{.347}\text{kcal}\text{.}$