Chapter 10, Problem 3SP

(a)

To determine

The amount of heat required to raise the ice to $0°\text{C}$ and completely melt the ice.

Answer to Problem 3SP

The heat required to raise the ice to $0°\text{C}$ is $13,500\text{cal}$

Explanation of Solution

Given info: The mass of ice is $150\text{g}$ at initial temperature $-20°\text{C}$. 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 ice has an initial temperature of $-20°C$. First of all ice has to raise the temperature to $0°\text{C}$ and then melt completely. Specific heat capacity helps in increasing the temperature and latent heat helps in melting the ice.

Write the expression for heat in terms of specific heat capacity of ice to raise the temperature to $0°\text{C}$.

$Q=mc\Delta T$ (1)

Here,

$Q$ is the heat energy required to raise the temperature of ice

$m$ is the mass of ice

$c$ 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 of ice

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

Substitute equation (2) in (1)

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

Substitute $150\text{g}$ for $m$, $0.5\text{cal/g}\cdot °\text{C}$ for $c$, $-20°\text{C}$ for $T_1$ and $0°\text{C}$ for $T_2$ in equation (3) to get the value of $Q$.

$\begin{array}{c}Q=150\text{g}\times 0.5\text{cal/g}\cdot °\text{C}\left(0°\text{C}-\left(-20°\text{C}\right)\right)\\ =1500\text{cal}\end{array}$

Write the equation to find the latent heat of fusion required to melt the ice.

$Q_l=mL$ (4)

Here,

$Q_l$ is the latent heat of fusion required to melt the ice.

$m$ is the mass of the ice

$L$ is the latent heat of fusion of ice

Substitute $150\text{g}$ for $m$ and $80\text{cal}/\text{g}$ for $L$ in equation ($) to get $Q_l$.

$\begin{array}{c}{Q}_l=150\text{g}\times 80\text{cal}/\text{g}\\ =12000\text{cal}\end{array}$

Thus the total heat required to raise the temperature of ice from $-20°\text{C}$ to $0°\text{C}$ and then to melt it completely is equal to the sum of specific heat and the latent heat of fusion.

$\begin{array}{c}{Q}_t=12000\text{cal+1500}\text{cal}\\ \text{=13,500}\text{cal}\end{array}$

Conclusion:

Therefore, the heat required to raise the ice to $0°\text{C}$ and then to melt it completely is $\text{13,500}\text{cal}$.

(b)

To determine

The additional heat required to heat the water.

Answer to Problem 3SP

The additional heat required to heat the water is $3750\text{cal}$

Explanation of Solution

Here the temperature of water has to be raised from ice temperature to $25°\text{C}$.

Write the expression to find the specific heat of water needed to be supplied to heat the water.

$Q=m{c}_W\left(T_2-T_1\right)$ (1)

Here,

$Q$ is the heat required to heat water

$m$ is the mass of water

$c_W$ is the specific heat capacity of water

$T_1$ is the initial temperature

$T_2$ is the final temperature

Substitute $150\text{g}$ for $m$ and $1\text{cal}/\text{g°C}$ for $c_W$ , $25°\text{C}$ for $T_2$ and $0°\text{C}$ for $T_1$ in equation (1) to get $Q$.

$\begin{array}{c}Q=150\text{g}\times 1\text{cal}/\text{g°C}\left(25°\text{C}-0°\text{C}\right)\\ =3750\text{cal}\end{array}$

Conclusion:

The heat required to heat the water is $3750\text{cal}$.

(c)

To determine

The total heat required to convert the $80\text{ g}$ of ice at $-20°\text{C}$ raise the temperature of water to $25°\text{C}$.

Answer to Problem 3SP

The heat required to raise the temperature of water to $25°\text{C}$ is $\text{17,250 cal}$.

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 of ice to raise the temperature to $0°\text{C}$.

$Q=mc\Delta T$ (1)

Here,

$Q$ is the heat energy required to raise the temperature of ice

$m$ is the mass of ice

$c$ 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 of ice

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

Substitute equation (2) in (1)

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

Substitute $150\text{g}$ for $m$, $0.5\text{cal/g}\cdot °\text{C}$ for $c$, $-20°\text{C}$ for $T_1$ and $0°\text{C}$ for $T_2$ in equation (3) to get the value of $Q$.

$\begin{array}{c}Q=150\text{g}\times 0.5\text{cal/g}\cdot °\text{C}\left(0°\text{C}-\left(-20°\text{C}\right)\right)\\ =1500\text{cal}\end{array}$

Write the equation to find the latent heat of fusion required to melt the ice.

$Q_l=mL$ (4)

Here,

$Q_l$ is the latent heat of fusion required to melt the ice.

$m$ is the mass of the ice

$L$ is the latent heat of fusion of ice

Substitute $150\text{g}$ for $m$ and $80\text{cal}/\text{g}$ for $L$ in equation ($) to get $Q_l$.

$\begin{array}{c}{Q}_l=150\text{g}\times 80\text{cal}/\text{g}\\ =12000\text{cal}\end{array}$

Thus the total heat required to raise the temperature of ice from $-20°\text{C}$ to $0°\text{C}$ and then to melt it completely is equal to the sum of specific heat and the latent heat of fusion.

$\begin{array}{c}{Q}_t=12000\text{cal+1500}\text{cal}\\ \text{=13500}\text{cal}\end{array}$

Write the expression for heat in terms of specific heat capacity required to raise the temperature of water to $25°\text{C}$

$Q=mc\Delta T$ (5)

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

$\Delta \text{T}\text{=}{\text{T}}_2-{\text{T}}_1$ (6)

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 (6) in (5)

$Q=mc\left({\text{T}}_2-{\text{T}}_1\right)$ (7)

Substitute $150\text{g}$ for $m$, $\text{1 cal/g}\cdot °\text{C}$ for $c$, $0°\text{C}$ for $T_1$ and $25°\text{C}$ for $T_2$ in equation (7)

$\begin{array}{c}Q=150\text{g}\times 1\text{cal/g}\cdot °\text{C}\left(25°\text{C}-0°\text{C}\right)\\ =3750\text{cal}\end{array}$

Therefore the heat required to raise the temperature of ice from $-20°\text{C}$ to $25°\text{C}$ is the sum of

Specific heat and latent heat of fusion.

$\begin{array}{c}{Q}_t=13500\text{cal+3750}\text{cal}\\ \text{=17,250}\text{cal}\end{array}$

Conclusion:

The heat required to raise the temperature of water to $25°\text{C}$ is $17,250\text{cal}$.

(d)

To determine

Whether we can just add the heat energies to find the heat needed to raise the temperature from $-20°\text{C}$ to $45°\text{C}$.

Answer to Problem 3SP

No we cannot add the heat energies straight to find the total heat as the value of specific heat of ice is not the same as that for water.

Explanation of Solution

The heat energy required to raise the temperature of ice from a temperature of $-20°\text{C}$ to $0°\text{C}$ is to be found first. After that the latent heat required to melt the ice with a zero centigrade temperature is to be found using the equation $m{L}_f$. Then the heat energy needed to raise the temperature of melted ice from a temperature of $0°\text{C}$ to $45°\text{C}$ is to be found. All these heat energy values are different since the value of specific heat, latent heat of fusion are different in each case. Therefore one cannot find the answer directly. It should be found considering each and every state of phase change separately.

Conclusion:

No we cannot add the heat energies straight to find the total heat as the value of specific heat of ice is not the same as that for water.