Chapter 10, Problem 3SP

(a)

To determine

The heat required to raise the ice to $0°\text{C}$.

Answer to Problem 3SP

The heat required to raise the ice to $0°\text{C}$ is $1540\text{cal}$.

Explanation of Solution

Given info: The mass of ice is $140\text{g}$ at initial temperature $-22°\text{C}$. The specific heat capacity of ice is $0.5\text{cal/g}\cdot °\text{C}$ and the specific heat capacity of water is $1\text{cal/g}\cdot °\text{C}$.

Write the expression for heat in terms of specific heat capacity

$Q=mc\Delta T$ (1)

Here,

$Q$ is the heat

$m$ is the mass

$c$ is the specific heat capacity of ice

$\Delta T$ is the change in temperature

Write the expression for change in temperature

$\Delta \text{T}\text{=}{\text{T}}_2-{\text{T}}_1$ (2)

Here,

$\Delta \text{T}$ is the change in temperature

${\text{T}}_1$ is the initial temperature

${\text{T}}_2$ is the final temperature

Substitute equation (2) in (1)

$Q=mc\left({\text{T}}_2-{\text{T}}_1\right)$ (3)

Substitute $140\text{g}$ for $m$, $0.5\text{cal/g}\cdot °\text{C}$ for $c$, $-22°\text{C}$ for $T_1$ and $0°\text{C}$ for $T_2$ in equation (3)

$\begin{array}{c}Q=140\text{g}\times 0.5\text{cal/g}\cdot °\text{C}\left(0°\text{C}-\left(-22°\text{C}\right)\right)\\ =1540\text{cal}\end{array}$

Conclusion:

The heat required to raise the ice to $0°\text{C}$ is $1540\text{cal}$.

(b)

To determine

The additional heat required to completely melt ice.

Answer to Problem 3SP

The additional heat required to completely melt ice is $11.2\text{kcal}$.

Explanation of Solution

Given info: The mass of ice is $140\text{g}$ at initial temperature $-22°\text{C}$. The specific heat capacity of ice is $0.5\text{cal/g}\cdot °\text{C}$ and the specific heat capacity of water is $1\text{cal/g}\cdot °\text{C}$.

Write the expression for heat in terms of latent heat capacity

$Q=m{L}_f$ (1)

Here,

$Q$ is the heat required tot completely melt ice

$m$ is the mass

$L_f$ is the latent heat capacity

Substitute $140\text{g}$ for $m$ and $80\text{cal/g}$ for $L_f$

$\begin{array}{c}Q=140\text{g}\times 80\text{cal/g}\\ \text{=11200}\text{cal}\\ =11.2\text{kcal}\end{array}$

Conclusion:

The additional heat required to completely melt ice is $11.2\text{kcal}$.

(c)

To determine

The heat required to raise the temperature of water to $27°\text{C}$.

Answer to Problem 3SP

The heat required to raise the temperature of water to $27°\text{C}$ is $3780\text{cal}$.

Explanation of Solution

Given info: The mass of ice is $140\text{g}$ at initial temperature $-22°\text{C}$. The specific heat capacity of ice is $0.5\text{cal/g}\cdot °\text{C}$ and the specific heat capacity of water is $1\text{cal/g}\cdot °\text{C}$.

Write the expression for heat in terms of specific heat capacity

$Q=mc\Delta T$ (1)

Here,

$Q$ is the heat

$m$ is the mass

$c$ is the specific heat capacity of ice

$\Delta T$ is the change in temperature

Write the expression for change in temperature

$\Delta \text{T}\text{=}{\text{T}}_2-{\text{T}}_1$ (2)

Here,

$\Delta \text{T}$ is the change in temperature

${\text{T}}_1$ is the initial temperature

${\text{T}}_2$ is the final temperature

Substitute equation (2) in (1)

$Q=mc\left({\text{T}}_2-{\text{T}}_1\right)$ (3)

Substitute $140\text{g}$ for $m$, $\text{1 cal/g}\cdot °\text{C}$ for $c$, $0°\text{C}$ for $T_1$ and $27°\text{C}$ for $T_2$ in equation (3)

$\begin{array}{c}Q=140\text{g}\times 1\text{cal/g}\cdot °\text{C}\left(27°\text{C}-0°\text{C}\right)\\ =3780\text{cal}\end{array}$

Conclusion:

The heat required to raise the temperature of water to $27°\text{C}$ is $3780\text{cal}$.

(d)

To determine

The total heat required to convert ice at $-22°\text{C}$ to water at $27°\text{C}$.

Answer to Problem 3SP

The total heat required to convert ice at $-22°\text{C}$ to water at $27°\text{C}$ is $16520\text{cal}$.

Explanation of Solution

Given info: The mass of ice is $140\text{g}$ at initial temperature $-22°\text{C}$. The specific heat capacity of ice is $0.5\text{cal/g}\cdot °\text{C}$ and the specific heat capacity of water is $1\text{cal/g}\cdot °\text{C}$.

The total heat will be the sum of the heat required to rise the temperature of ice from $-22°\text{C}$ to $0°\text{C}$, and heat required to completely melt ice at $0°\text{C}$ and the heat required to rise the temperature water obtained by melting ice to $27°\text{C}$

$Q_{\text{total}}={\left(mc\left(T_1-T_2\right)\right)}_{\text{ice}}+m{L}_f+{\left(mc\left(T_1-T_2\right)\right)}_{water}$

Substitute $1540\text{cal}$ for the first term, $11200\text{cal}$ for the second term, $3780\text{cal}$ for the third term

$\begin{array}{c}{Q}_{\text{total}}=1540\text{cal+11200}\text{cal+3780}\text{cal}\\ \text{=16520}\text{cal}\\ \text{=16}\text{.52}\text{kcal}\end{array}$

Conclusion:

The total heat required to convert ice at $-22°\text{C}$ to water at $27°\text{C}$ is $16520\text{cal}$.

(e)

To determine

Which takes more energy in the processes mentioned in part a, b, c.

Answer to Problem 3SP

Melting of ice completely takes high energy of $11.2\text{kcal}$.

Explanation of Solution

Given info: The mass of ice is $140\text{g}$ at initial temperature $-22°\text{C}$. The specific heat capacity of ice is $0.5\text{cal/g}\cdot °\text{C}$ and the specific heat capacity of water is $1\text{cal/g}\cdot °\text{C}$.

The heat required to raise the temperature from $-22°\text{C}$ to $0°\text{C}$ is $1540\text{cal}$, heat required to melt ice completely is $11.2\text{kcal}$ and heat required to heat water to $27°\text{C}$ is $3780\text{cal}$.

In which more heat is consumed for the intermediate process, which is melting of ice.

Conclusion:

Melting of ice completely takes high energy of $11.2\text{kcal}$.

(f)

To determine

To explain whether we can find this total heat simply by computing how much heat is required to melt the ice and adding the amount of heat required to raise the temperature of $140 \text{g}$ water by $49°\text{C}$ .

Answer to Problem 3SP

No, before melting the temperature of ice should from $-22°\text{C}$ to $0°\text{C}$. it is also needed to include to complete the calculation.

Explanation of Solution

Given info: The mass of ice is $140\text{g}$ at initial temperature $-22°\text{C}$. The specific heat capacity of ice is $0.5\text{cal/g}\cdot °\text{C}$ and the specific heat capacity of water is $1\text{cal/g}\cdot °\text{C}$.

Three processes are takes place in the conversion of ice to water. Fist ice should be raised from its initial temperature to $0°\text{C}$ and then to completely melt ice to convert to water finally to raise the temperature of water from $0°\text{C}$ to $27°\text{C}$.

In different stages, the heat capacity will be different. In the intermediate stage during phase change it is required to consider the latent heat of fusion.

Conclusion:

No, before melting the temperature of ice should from $-22°\text{C}$ to $0°\text{C}$. it is also needed to include to complete the calculation.