Question

Three different materials of identical mass are placed one at a time in a special freezer that can extract energy from a material at a certain constant rate. During the cooling process, each material begins in the liquid state and ends in the solid state; Fig. 18-28 shows the temperature Tversus time t. (a) For material 1, is the specific heat for the liquid state greater than or less than that for the solid state? Rank the materials according to (b) freezing point temperature, (c) specific heat in the liquid state, (d) specific heat in the solid state, and (e) heat of fusion, all greatest first.

Short answer

1. For material, the specific heat for the liquid state is greater than that for the solid state.
2. The rank of the material according to freezing point temperature is$\mathrm{material}1>\mathrm{material}2>\mathrm{material}3$.
3. The rank of the material according to specific heat in the liquid state is$\mathrm{material}1>\mathrm{material}3>\mathrm{material}2$.
4. The rank of the material according to specific heat in the solid state is$\mathrm{material}1>\mathrm{material}2>\mathrm{material}3$.
5. The rank of the material according to heat of fusion is $\mathrm{material}2>\mathrm{material}3>\mathrm{material}1$.

Step-by-step solution

The given data

1. All materials have the same mass .
2. A special freezer extracts energy from the materials at a constant rate.
3. The Figure 18-28 is given.

Understanding the concept of thermal expansion

When an object's temperature changes, it expands and grows larger, a process known as thermal expansion. We can write the slope of the given graph by rearranging the formula for the heat transfer. From this, we can write the specific heat in terms of the slope. Observing the slope of the graph, we can determine whether the specific heat for the liquid state is greater than or less than that for the solid-state. Also observing the graph, we can rank the materials according to the freezing point temperature, specific heat in the liquid state, specific heat in the solid-state, and heat of fusion.

Formulae:

The energy transferred by the body as heat to the surroundings,$\mathrm{Q}=\mathrm{mc}∆\mathrm{T}$ …(i)

The total energy transformed due to latent heat of fusion,$\mathrm{Q}={\mathrm{L}}_{\mathrm{F}}\mathrm{m}$ …(ii)

Where,

m is mass, c is specific heat , $∆\mathrm{T}$is change in temperature and ${\mathrm{L}}_{\mathrm{F}}$is the heat of fusion

(a) Calculation of the specific heat for material 1

From equation (i), the ratio of heat transferred by time difference is given by:

$\frac{\mathrm{Q}}{\mathrm{t}}=\frac{\mathrm{cm}∆\mathrm{T}}{\mathrm{t}}$ …(iii)

But, three different materials are placed in a special freezer that can extract energy from a material at a certain constant rate. That is,data-custom-editor="chemistry" $\frac{\mathrm{Q}}{\mathrm{t}}=\mathrm{constant}$

Thus, using this condition in equation (iii), we get that

data-custom-editor="chemistry" $\frac{\mathrm{cm}∆\mathrm{T}}{\mathrm{t}}=\mathrm{constant}$

But all materials havethesame mass m. Thus, the above condition becomes:

data-custom-editor="chemistry" $\mathrm{c}\times \frac{∆\mathrm{T}}{\mathrm{t}}=\mathrm{constant}$

where,data-custom-editor="chemistry" $\frac{∆\mathrm{T}}{\mathrm{t}}=\mathrm{slope}\mathrm{of}\mathrm{T}\mathrm{vs}\mathrm{t}\mathrm{graph}$

Thus, the value of specific heat becomes:

$\mathrm{c}=\frac{1}{\mathrm{slope}}$

In the Figure 18-28, we can see that high temperature portion of the graph representsthe liquid state and lower temperature portion represents the solid state of the materials.

In the Figure 18-28, we can see that for material 1, data-custom-editor="chemistry" $\frac{∆\mathrm{T}}{\mathrm{t}}=\mathrm{slope}\mathrm{of}\mathrm{T}\mathrm{vs}\mathrm{t}\mathrm{graph}$is smaller for the liquid state than that of the solid state.

Hence, the specific heat c is larger for the liquid state than that for the solid state.

(b) Calculation of the rank of the material according to freezing point temperature

In the Figure 18-28, the horizontal lines between the liquid state and solid state in the $\mathrm{T}\mathrm{vs}\mathrm{t}$graph gives the freezing point temperature of corresponding material. Therefore, in the Figure 18-28, we can see that the horizontal line between the liquid state and solid state for material 1 is above that of material 2 and also the horizontal line between the liquid state and solid state for material 2 is above that of material 3.

Thus, we get the freezing point as: ${\mathrm{T}}_1>{\mathrm{T}}_2>{\mathrm{T}}_3$

Hence, the rank according to the freezing temperatures as $\mathrm{material}1>\mathrm{material}2>\mathrm{material}3$, where, ${\mathrm{T}}_1,{\mathrm{T}}_2$, and T₃ are the freezing temperatures of the materials, 1,2 and 3 respectively.

(c) Calculation of the rank of the material according to specific heat of the liquid state

From calculations of part (a), we can see that

$\mathrm{c}\times \frac{∆\mathrm{T}}{\mathrm{t}}=\mathrm{constant}$

where,$\frac{∆\mathrm{T}}{\mathrm{t}}=\mathrm{slope}\mathrm{of}\mathrm{T}\mathrm{vs}\mathrm{t}\mathrm{graph}$

In the Figure 18-28, we can see that high temperature portion of the graph representsthe liquid state and lower temperature portion represents the solid state of the materials.

In the Figure 18-28, we can see that for material 1, $\frac{∆\mathrm{T}}{\mathrm{t}}=\mathrm{slope}\mathrm{of}\mathrm{T}\mathrm{vs}\mathrm{t}\mathrm{graph}$is smaller for the liquid state than that for material 3.

Hence, the specific heat for the liquid state of material 1 is larger than that of material 3. Also, for material 3, $\frac{∆\mathrm{T}}{\mathrm{t}}=\mathrm{slope}\mathrm{of}\mathrm{T}\mathrm{vs}\mathrm{t}\mathrm{graph}$is smaller for the liquid state than that for material 2. Thus, the specific heat for the liquid state of material 3 is larger than that of material 2.

Thus, we get the specific heat as: ${\mathrm{c}}_{1,\mathrm{l}}>{\mathrm{c}}_{3,\mathrm{l}}>{\mathrm{c}}_{2,\mathrm{l}}$

Hence, the rank of the material according to the specific heat of the liquid state as $\mathrm{material}1>\mathrm{material}3>\mathrm{material}2$, where, ${\mathrm{c}}_{1,\mathrm{l}},{\mathrm{c}}_{2,\mathrm{l}}$, and ${\mathrm{c}}_{3,\mathrm{l}}$are the specific heats for the liquid state of the material 1,2, and 3 respectively.

(d) Calculation of the rank of the material according to specific heat of the solid state

From calculations of part (a), we can see that

$\mathrm{c}\times \frac{∆\mathrm{T}}{\mathrm{t}}=\mathrm{constant}$

where,$\frac{∆\mathrm{T}}{\mathrm{t}}=\mathrm{slope}\mathrm{of}\mathrm{T}\mathrm{vs}\mathrm{t}\mathrm{graph}$

In the Figure 18-28, we can see that high temperature portion of the graph representsthe liquid state and lower temperature portion represents the solid state of the materials.

In the Figure 18-28, we can see that for material 1, $\frac{∆\mathrm{T}}{\mathrm{t}}=\mathrm{slope}\mathrm{of}\mathrm{T}\mathrm{vs}\mathrm{t}\mathrm{graph}$is smaller for the solid state than that for material 2. Hence, the specific heat for the solid state of material 1 is larger than that of material 2.

Also, for material 2, $\frac{∆\mathrm{T}}{\mathrm{t}}=\mathrm{slope}\mathrm{of}\mathrm{T}\mathrm{vs}\mathrm{t}\mathrm{graph}$is smaller for the solid state than that for material 3. Hence, the specific heat for the solid state of material 2 is larger than that of material 3. Thus, we get the specific heat as: ${\mathrm{c}}_{1,\mathrm{s}}>{\mathrm{c}}_{2,\mathrm{s}}>{\mathrm{c}}_{3,\mathrm{s}}$, where, ${\mathrm{c}}_{1,\mathrm{s}},{\mathrm{c}}_{2,\mathrm{s}}$, and ${\mathrm{c}}_{3,\mathrm{s}}$are the specific heats for the solid state of the material, 1,2 and 3 respectively.

Hence, the rank of the material according to their specific heat of the solid state is $\mathrm{material}1>\mathrm{material}2>\mathrm{material}3$.

(e) Calculation of the rank of the material according to the heat of fusion

The heat of fusion of the material is given using equation (ii) as follows:

${\mathrm{L}}_{\mathrm{F}}=\frac{\mathrm{Q}}{\mathrm{m}}$

In the Figure 18-28, we can see that there is a horizontal line in the graph for each material. This horizontal line gives the freezing point of the corresponding material.

But at the freezing point, the temperature T is constant, but time t changes; heat Q is released to change the state of the material from liquid to solid, and that heat per unit mass of the material is called the heat of fusion data-custom-editor="chemistry" ${\mathrm{L}}_{\mathrm{F}}$. If all materials have identical mass m then the heat of fusion data-custom-editor="chemistry" ${\mathrm{L}}_{\mathrm{F}}$of the material depends on the heat Q which is released to change the state of the material from liquid to solid.

In the Figure 18-28, we can see that the length of the horizontal line for material 2 is higher than that for material 3, and also the length of the horizontal line for material 3 is higher than that for material 1. It means in material 2, the heat Q released at a constant rate for time t is greater than that for material 3 and also in material 3, the heat Q released at a constant rate for time t is greater than that for material 1. Thus, we get the heat of fusion as: role="math" localid="1661864766315" ${\mathrm{L}}_{\mathrm{F},2}>{\mathrm{L}}_{\mathrm{F},3}>{\mathrm{L}}_{\mathrm{F},1}$ where, L_F1 , L_F2 and L_F3 are the heats of fusion for material 1, 2 and 3 respectively.

Hence, the rank of the materials according to the heat of fusion is data-custom-editor="chemistry" $\mathrm{material}2>\mathrm{material}3>\mathrm{material}1$.