Chapter 7, Problem 93P

Use the following values to answer each part. The specific heat of water is $\text{1}\text{.00 cal/(g}\cdot °\text{C);}$ the heat of fusion of water is $\text{79-7 cal/g;}$ and the heat of vaporization of wateris $\text{540 cal/g}\text{.}$

1. How much energy (in calories) is needed to melt 45 g of ice at $\text{0}\text{.0 }°\text{C}$ and warm it to $\text{55 }°\text{C}$?
2. How much energy (in calories) is released when 45 g of water at $\text{55 }°\text{C}$ is cooled to $\text{0}\text{.0 }°\text{C}$ , and frozen to solid ice at $\text{0}\text{.0 }°\text{C}$?
3. How much energy (in kilocalories) is released when 35 g of steam at $\text{100 }°\text{C}$ is condensed to water, the water is cooled to $\text{0}\text{.0 }°\text{C}$ , and the water is frozen to solid ice at $\text{0}\text{.0 }°\text{C}$?

Interpretation Introduction

(a)

Interpretation:

The amount of heat required to melt 45 g of ice at 0 degree Celsius and warm to 55 degree Celsius needs to be determined.

Concept Introduction:

The amount of heat released or absorbed can be calculated as follows:

  $Q=m\times C\times \Delta T$

Here,

Q = amount of energy

m = mass

C = specific heat

  $\Delta T$ = change in temperature

Answer to Problem 93P

The total energy which is required to melt 45g of ice at $0.0°C$ and warm it to $55°C$ is 6061.5 cal.

Explanation of Solution

The change in temperature can be calculated by the following formula −

  $Q=m\times C\times \Delta T$

Specific heat of water, C = $1.00cal/g.°C$

  $\Delta T$ = $\left(55°C-0°C\right)$

Mass of ice, m = 45

Put the above values in above formula −

  $Q=45g\times 1.00cal/g.°C\times \left(55°C-0°C\right)$

  $Q=2475cal$

Now, energy is absorbed, therefore heat of fusion of water = $79.\text{7 cal/g}$

The amount of heat for the fusion of 45 g of freezing water is calculated as −

  $\text{45g×}\frac{\text{79}\text{.7cal}}{\text{1}\text{.0g}}\text{=3586}\text{.5cal}$

Therefore, total amount of energy is calculated as −

  $=\text{3586}\text{.5 cal+2475 cal}$

Total amount of energy = 6061.5 cal

Interpretation Introduction

(b)

Interpretation:

The amount of energy liberated needs to be determined, when 45g of water at $55°\text{C}$ is cooled to $0.0°\text{C}$ and frozen to solid ice at $0.0°\text{C}$.

Concept Introduction:

The amount of heat released or absorbed can be calculated as follows:

  $Q=m\times C\times \Delta T$

Here,

Q = amount of energy

m = mass

C = specific heat

  $\Delta T$ = change in temperature

Answer to Problem 93P

The amount of energy released when 45g of water at $55°C$ is cooled to $0.0°C$ and frozen to solid ice at $0.0°C$ is $-6061.\text{5 cal}$.

Explanation of Solution

Energy is releasing, therefore heat of fusion of water = $-79.\text{7cal/g}$.

The amount of heat for the fusion of 45g of freezing water is calculated as −

  $\text{45g×}\frac{\text{-79}\text{.7cal}}{\text{1}\text{.0g}}\text{=-3586}\text{.5cal}$

The change in temperature can be calculated by the following formula −

  $Q=m\times C\times \Delta T$

Specific heat of water, C = $\text{1}\text{.00cal/g}\text{.°C}$

  $\Delta T$ = $\left(55°C-0°C\right)$

Mass of ice, m = 45g

Put the above values in above formula −

  $Q=\text{45g×1}\text{.00cal/g}\text{.°C×}\left(\text{0°C-55°C}\right)$

  $Q=-2\text{475 cal}$

Therefore, total amount of energy is calculated as −

  $=-3586.\text{5 cal}+(-247\text{5) cal}$

Total amount of energy = -6061.5 cal

Interpretation Introduction

(c)

Interpretation:

The amount of energy liberated is to be determined when 35g of steam at $100°\text{C}$ is condensed to water, the water is cooled to $0.0°C$ and frozen to solid ice at $0.0°C$.

Concept Introduction:

The amount of heat released or absorbed can be calculated as follows:

  $Q=m\times C\times \Delta T$

Here,

Q = amount of energy

m = mass

C = specific heat

  $\Delta T$ = change in temperature

Answer to Problem 93P

The total energy released when 35g of steam at $100°C$ is condensed to water, the water is cooled to $0.0°C$ and frozen to solid ice at $0.0°C$ is $-25.\text{1 kcal}$.

Explanation of Solution

Heat of vaporization = 540cal/g

It is condensation, then heat of condensation = $-54\text{0cal/g}$

The amount of heat for 35 g of water can be calculated as −

  $\text{35g×}\frac{\text{-540cal}}{\text{1}\text{.0g}}\text{=-18900cal}$

The change in temperature can be calculated by the following formula −

  $Q=m\times C\times \Delta T$

Specific heat of water, C = $1.0\text{0cal/g}\text{.°C}$

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

Mass of ice, m = 35g

Put the above values in above formula −

  $\text{Q=35g×1}\text{.00cal/g}\text{.°C×}\left(\text{0°C-100°C}\right)$

  $Q=-\text{3500 cal}$

As given is releasing, and then heat of fusion for 35 g freezing water is calculated as −

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

  $\text{35g×}\frac{\text{-79}\text{.7cal}}{\text{1}\text{.0g}}\text{=-2789}\text{.5cal}$

Therefore, total amount of energy is calculated as −

  $=-1890\text{0 cal}+(-350\text{0 cal})+\left(-278\text{9}\text{.5 cal}\right)$

Total amount of energy = -25189.5 cal

The conversion of unit −

1kcal = 1000cal

Therefore,

  $=\frac{\text{-25189}\text{.5 cal}}{\text{1000}}\text{=-25}\text{.1 kcal}$

Total amount of energy = -25 kcal