Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

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Short Answer

A 0.400 kg sample is placed in a cooling apparatus that removes energy as heat at a constant rate. Figure 18-32 gives the temperature T of the sample versus time t; the horizontal scale is set by t_s=80.0 min. The sample freezes during the energy removal. The specific heat of the sample in its initial liquid phase is 300 J/kgK . (a) What is the sample’s heat of fusion and (b) What is its specific heat in the frozen phase?

The sample’s heat of fusion is 68 kJ/kg
The specific heat of the sample in the frozen phase is $2.3 kJ / {kg}^{°} C$

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Identification of given data

The mass of the sample is $m = 0.400 kg$
The time of the sample is $t_{s} = 80.0 \min$
The specific heat of the sample is $c = 3000 J / kg . K$

Step 2: Significance of specific heat

The specific heat capacity of water is the amount of heat required to change the temperature of a unit mass of water by one degree.

We can use the concept of specific heat of the water and find the energy

transferred during this time. We can use the expression of the power consumed by the system and then use the expression of the heat of fusion and the specific heat of the sample in the frozen phase.

Formulae:

The heat energy required by a body, $Q = mc ∆ T$ …(i)

Here, m is mass, c is specific heat capacity, $∆ T$ is change in temperature, Q is required heat energy.

The heat energy released or absorbed by the body, $Q = L_{F} m$ …(ii)

Here, L is specific latent heat of fusion.

The power exerted through the heat by a body, $P = \frac{Q}{t}$ …(iii)

Here, P is the power, Q is required heat energy, t is the time.

Step 3: (a) Determining the sample’s heat of fusion

According to the figure, the temperature decreases from $T_{i} = 300^{°} C$ to $T_{f} = 270^{°} C$ within time, $t = 40.0 \min$ .

The energy transferred during this time using equation (i) is given by:

$\begin{array}{rcl} Q & = & 3000 J / kg . {}^{\circ}C \times 0.40 kg \times |270^{\circ} C - 300^{\circ} C| \\ = & 36000 J \end{array}$

The transformation rate of the heat can be given using equation (ii) as:

$\begin{array}{rcl} P & = & \frac{36000 J}{40 \min} \\ = & 900 \frac{J}{\min} \\ = & \frac{900 J}{60 s} \\ = & 15 W \end{array}$

During next $t = 30.0 \min$ , the heat transferred by the body using equation (iii) is given as:

$\begin{array}{rcl} Q & = & 900 \frac{J}{\min} \times 30 \min \\ = & 27000 J \end{array}$

The phase changes from liquid to the solid. Then, the heat of transformation involved in this phase change is called heat of fusion. The specific latent heat of fusion using equation (ii) can be given as:

$\begin{array}{rcl} L_{F} & = & \frac{Q}{m} \\ = & \frac{27000 J}{0.40 kg} \\ = & 67500 J / kg \\ = & 68 kJ / kg \end{array}$

Hence, the value of the specific heat of fusion is $68 kJ / kg$ .

Step 4: (b) Determining the specific heat of the sample in the frozen phase

During next $t = 20.0 \min$ , the sample is solid and change of temperature is $∆ T = 30^{\circ} C$ . The expression for the specific heat of the sample using equation (i) is given as:

$\begin{array}{rcl} c & = & \frac{Q}{m |∆ T|} \\ = & \frac{27000 J}{0.40 kg \times 30^{\circ} C} \\ = & 2250 \frac{J}{kg . {}^{\circ}C} \\ \approx & 2.3 \frac{kJ}{kg . {}^{\circ}C} \end{array}$

Hence, the value of the specific heat is $2.3 \frac{kJ}{kg . {}^{\circ}C}$

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Fundamentals Of Physics

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Short Answer

Step by Step Solution

Step 1: Identification of given data

Step 2: Significance of specific heat

Step 3: (a) Determining the sample’s heat of fusion

Step 4: (b) Determining the specific heat of the sample in the frozen phase

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Fundamentals Of Physics

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Short Answer

Step by Step Solution

Step 1: Identification of given data

Step 2: Significance of specific heat

Step 3: (a) Determining the sample’s heat of fusion

Step 4: (b) Determining the specific heat of the sample in the frozen phase

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