Chapter 2, Problem 2.1P

(a)

Interpretation Introduction

Interpretation:

To determine the amount of work done on the water.

Concept Introduction:

This is a case of steady flow process as we assume that the fluid flowing is incompressible and has a constant density throughout the flow. Thus, use the concept of calorimetry to find out the final temperature of water.

Answer to Problem 2.1P

The amount of work done on water is $0.175\text{kJ}$ .

Explanation of Solution

Given information:

Total mass, $M_{wt}=35\text{ kg}$

Acceleration due to gravity, $g=9.8{\text{ ms}}^{\text{-2}}$

Height, $\Delta z=5\text{ m}$

Specific heat of water, $C_P=4.18\text{ kJ}\cdot {\text{kg}}^{\text{-1,}}{℃}^{\text{-1}}.$

Mass of water, ${\text{M}}_{{\text{H}}_{\text{2}}\text{O}}=25\text{ kg}$

Initial temperature, ${\text{T}}_{\text{1}}\text{=20}\text{°C}$

The amount of work done on water is given by,

$\begin{array}{c}W=M_{wt}\times \Delta z\\ =35\times 5\\ =175\text{ J}\\ =0.175\text{ kJ}\end{array}$

Therefore, the amount of work done on water is $0.175\text{kJ}$ .

(b)

Interpretation Introduction

Interpretation:

To determine the internal energy change of the water.

Concept Introduction:

Internal Energy - The energy which is present within the system itself. It eliminates kinetic energy causes due to motion of the system and the potential energy of the system.

Answer to Problem 2.1P

The internal energy change of water is $1.715\text{ kJ}$ .

Explanation of Solution

Given Data:

Total mass, $M_{wt}=35\text{ kg}$

Acceleration due to gravity, $g=9.8{\text{ ms}}^{\text{-2}}$

Height, $\Delta z=5\text{ m}$

Specific heat of water, $C_P=4.18\text{ kJ}\cdot {\text{kg}}^{\text{-1,}}{℃}^{\text{-1}}.$

Mass of water, $M_{H_{2}O}=25\text{ kg}$

Initial temperature, $T_1=20℃$

This is a case of steady flow process as we assume that the fluid flowing is incompressible and has a constant density throughout the flow. You also use the concept of calorimetry to find out the final temperature of water.

Calculation:

The internal energy change of water is given by,

$\begin{array}{c}{U}_{total}=\text{Work}\left(W\right)\\ =1.715\text{ kJ}\end{array}$

Therefore, the internal energy change of water is $1.715\text{ kJ}$ .

(c)

Interpretation Introduction

Interpretation:

To determine the final temperature of water, for which $C_P=4.18\text{ kJ}\cdot {\text{kg}}^{\text{-1,}}{℃}^{\text{-1}}.$

Concept Introduction:

This is a case of steady flow process as we assume that the fluid flowing is incompressible and has a constant density throughout the flow. You also use the concept of calorimetry to find out the final temperature of water.

Answer to Problem 2.1P

The final temperature of water is $20.016°\text{C}$ .

Explanation of Solution

Given Data:

Total mass, $M_{wt}=35\text{ kg}$

Acceleration due to gravity, $g=9.8{\text{ ms}}^{\text{-2}}$

Height, $\Delta z=5\text{ m}$

Specific heat of water, $C_P=4.18\text{ kJ}\cdot {\text{kg}}^{\text{-1}}{℃}^{\text{-1}}.$

Mass of water, $M_{H_{2}O}=25\text{ kg}$

Initial temperature, $T_1=20℃$

This is a case of steady flow process as we assume that the fluid flowing is incompressible and has a constant density throughout the flow.

Calculation:

To determine the final temperature of water, use the first law of thermodynamics as well as with calorimetry given by the equation, $dU+d\left(PV\right)=C_P.dT$

Since P is constant the equation can also be written as,

$M_{H_{2}O}.C_P.dT=M_{H_{2}O}.dU+M_{H_{2}O}PdV$

Take $C_P$ and $V$ as constant and integrate the equation,

$\begin{array}{c}{M}_{H_{2}O}{C}_P{\displaystyle \int dT}={\displaystyle \int dU}\\ M_{H_{2}O}{C}_P\left(T_2-T_1\right)=U_{\text{total}}\\ T_2=T_1+\frac{U_{\text{total}}}{M_{H_{2}O}.C_P}\\ =20+\frac{1.715}{25\times 4.18}\\ =20+0.016\\ =20.016℃\end{array}$

Therefore, the final temperature of water is $20.016℃$ .

(d)

Interpretation Introduction

Interpretation:

To determine the amount of heat that must be removed from the water to return it to its initial temperature.

Concept Introduction:

The specific heat is the quantity of energy that is needed to increase the temperature by $1°C$ .

Answer to Problem 2.1P

The amount of heat removed from the water to return to its initial temperature is $-1.715\text{ kJ}$ .

Explanation of Solution

Given Data:

Total mass, $M_{wt}=35\text{ kg}$

Acceleration due to gravity, $g=9.8{\text{ ms}}^{\text{-2}}$

Height, $\Delta z=5\text{ m}$

Specific heat of water, $C_P=4.18\text{ kJ}\cdot {\text{kg}}^{\text{-1,}}{℃}^{\text{-1}}.$

Mass of water, $M_{H_{2}O}=25\text{ kg}$

Initial temperature, $T_1=20℃$

This is a case of steady flow process as we assume that the fluid flowing is incompressible and has a constant density throughout the flow. You also use the concept of calorimetry to find out the final temperature of water.

Calculation:

For the process of restoration, i.e. water returns to its initial temperature, the change in internal energy is equal but opposite to that of the initial process.

$\begin{array}{c}{\text{Q= -ΔU}}_{\text{total}}\\ \text{=-1}\text{.715 kJ}\end{array}$

Therefore, the amount of heat removed from the water to return to its initial temperature is $-1.715\text{ kJ}$ .

(e)

Interpretation Introduction

Interpretation:

To determine the total energy change of the universe because of (1) the process of lowering the weight, (2) the process of cooling the water back to its initial temperature, and (3) both processes together.

Concept Introduction:

The total energy of the system is equal to the sum of all the energies.

Answer to Problem 2.1P

The total energy change in universe due to process $\left(1\right),\left(2\right)\text{and}\left(3\right)$ is Zero

Explanation of Solution

Given Data:

Total mass, $M_{wt}=35\text{ kg}$

Acceleration due to gravity, $g=9.8{\text{ ms}}^{\text{-2}}$

Height, $\Delta z=5\text{ m}$

Specific heat of water, $C_P=4.18\text{ kJ}\cdot {\text{kg}}^{\text{-1,}}{℃}^{\text{-1}}.$

Mass of water, $M_{H_{2}O}=25\text{ kg}$

Initial temperature, $T_1=20℃$

This is a case of steady flow process as we assume that the fluid flowing is incompressible and has a constant density throughout the flow. You also use the concept of calorimetry to find out the final temperature of water.

Calculation:

The change in internal energy in all the processes i.e. Process $\left(1\right),\left(2\right)\text{and}\left(3\right)$ is zero since no heat is lost to the environment.

Therefore, the total energy change in universe due to process $\left(1\right),\left(2\right)\text{and}\left(3\right)$ is Zero.