Chapter 4, Problem 4.75AP

(a)

Interpretation Introduction

Interpretation:

The amount of heat energy needed to melt $\text{45}\text{g}$ of ice 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 4.75AP

The amount of energy needed to melt ice is $2475\text{cal}\text{.}$

Explanation of Solution

The amount of energy needed to melt ice:

Given,

The mass of ice melt is $45\text{g}\text{.}$

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 ${\text{0}}^{\text{o}}\text{C}\text{.}$

The temperature at final $\left({\text{T}}_2\right)$ is ${55}^{\text{o}}\text{C}\text{.}$

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

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

  ${\text{ΔT=55}}^{\text{o}}\text{C}$

The amount of energy needed to melt ice can be given as,

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

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

  $\text{Heat}\text{=}\text{45}{\text{g×55}}^{\text{o}}\text{C}\text{×}\text{}\text{1}\text{.00}{\text{cal/g}}^{\text{o}}\text{C}$

  $\text{Heat}\text{=}\text{2475}\text{cal}$

The amount of energy needed to melt ice is $2475\text{cal}\text{.}$

(b)

Interpretation Introduction

Interpretation:

The amount of heat energy released when $\text{45}\text{g}$ of water is cooled has to be given.

Concept introduction:

Refer to part (a).

Answer to Problem 4.75AP

The amount of energy released when water is cooled is $-2475\text{cal}\text{.}$

The amount of heat energy released when $\text{45}\text{g}$ of water frozen is $3587\text{cal}\text{.}$

Explanation of Solution

The amount of heat energy released when $\text{45}\text{g}$ of water is cooled:

Given,

The mass of water is $45\text{g}\text{.}$

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 ${55}^{\text{o}}\text{C}\text{.}$

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

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

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

  $\text{ΔT=}{\text{-55}}^{\text{o}}\text{C}$

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

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

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

  $\text{Heat}\text{=}\text{45}\text{g×-}{\text{55}}^{\text{o}}\text{C}\text{×}\text{}\text{1}\text{.00}{\text{cal/g}}^{\text{o}}\text{C}$

  $\text{Heat}\text{=}-\text{2475}\text{cal}$

The amount of energy released when water is cooled is $-2475\text{cal}\text{.}$

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

The mass of water is $45\text{g}\text{.}$

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

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

The amount of energy needed to melt ice can be given as,

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

  $45\text{g×}\frac{\text{79}\text{.7}\text{cal}}{\text{1g}}\text{=3587}\text{calories}$

The amount of heat energy released when $\text{45}\text{g}$ of water is frozen is $3587\text{cal}\text{.}$

(c)

Interpretation Introduction

Interpretation:

The amount of energy released when $\text{35}\text{g}$ of steam condensed has to be given.

Concept introduction:

Refer to part (a).

Answer to Problem 4.75AP

The amount of energy released when water is condensed is $\text{-}\text{3}\text{.500}\text{kcal}\text{.}$

The amount of heat energy released when $35\text{g}$ of water is cooled is $\text{2}\text{.79}\text{kcal}\text{.}$

Explanation of Solution

The amount of energy when steam condensed:

Given,

The mass of steam is $35\text{g}\text{.}$

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

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

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

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 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,

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

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

  $\text{Heat}\text{=}35\text{g×-}{100}^{\text{o}}\text{C}\text{×}\text{}\text{1}\text{.00}{\text{cal/g}}^{\text{o}}\text{C}$

  $\text{Heat}\text{=}-3500\text{cal}$

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

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

  $\text{-3500}\cancel{\text{cal}}\text{×}\frac{\text{1}\text{kcal}}{\text{1000}\cancel{\text{cal}}}\text{=}\text{-3}\text{.500}\text{kcal}$

The amount of energy released when water is condensed is $\text{-}3.5\text{00}\text{Kcal}\text{.}$

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,

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

  $35\text{g×}\frac{\text{79}\text{.7}\text{cal}}{\text{1g}}\text{=3347}\text{calories}$

The amount of heat energy released when $35\text{g}$ of water is frozen is $2790\text{cal}\text{.}$

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

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

  $\text{2790}\cancel{\text{cal}}\text{×}\frac{\text{1}\text{kcal}}{\text{1000}\cancel{\text{cal}}}\text{=2}\text{.79}\text{kcal}$

The amount of heat energy released when $35\text{g}$ of water is frozen is $\text{2}\text{.79}\text{kcal}\text{.}$