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Physics

1st Edition

Walker

ISBN: 9780133256925

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Page 374: Lesson Check

Exercise 47

Solution 1

Solution 2

Step 1

1 of 2

An equilibrium means stability. Thus there will be no change in the phase if they are in equilibrium.

Thus if x number of particles enter a particular phase, the number of particles leaving the phase will also be x

Therefore, phase equilibrium is established when the number of net particles remain constant in the phase.

Result

2 of 2

Therefore, phase equilibrium is established when the number of net particles remain constant in the phase.

Step 1

1 of 1

If two phases are in equilibrium, then the number of molecule leaving one phase is equal to the number of molecule entering the same phase. So the mass of the phase will not change over time. Then we will know that the phase are in equilibrium.

Exercise 48

Step 1

1 of 2

Through evaporation, the particles of water with the highest kinetic energy leave the droplet. This leads to a decrease of the average kinetic energy, that is temperature of the droplet.

Since the droplet now has a smaller temperature than the human skin it is in contact with, it absorbs heat from the skin.

Hence the thermal energy of the body leaves it in the form of high speed water molecules.

Result

2 of 2

The droplet of sweat absorbs heat, needed to evaporate, from the skin and cools it.

Exercise 49

Step 1

1 of 2

The amount of thermal energy required to change the phase of one kg of any substance without changing the temperature is called latent heat.

Step 2

2 of 2

Latent heat does not change the temperature of the substance.

Exercise 50

Solution 1

Solution 2

Step 1

1 of 2

The force per area is called pressure.

Result

2 of 2

Pressure.

Step 1

1 of 2

Force per unit area is known as pressure. Pressure is the amount of force exerted on a given area. The symbol for pressure is $P$, and the pressure is produced by a force $F$ on an area $A$ is given by the following:

$$
text{pressure} = dfrac{text{force}}{text{area}}
$$

Therefore,

$$
p = dfrac{F}{A}
$$

SI units of pressure is $dfrac{text{N}}{text{m}^2}$.

Result

2 of 2

$$
Pressure = dfrac{F}{A}
$$

Exercise 51

Step 1

1 of 2

The pressure of a gas that is in equilibrium with its liquid phase is called the $textbf{equilibrium vapor pressure}$.

This is the pressure at which the same number of particles transition from the liquid phase to the gas phase as the number of particles that transition from the gas phase to the liquid phase each second. Hence dynamic equilibrium is reached.

Result

2 of 2

Equilibrium vapor pressure

Exercise 52

Solution 1

Solution 2

Step 1

1 of 2

(a) The water will be in liquid form because from the figure we can see that the boiling temperature of water in pressure cooker is $120^circ text{C}$, and the given temperature is less than the boiling point therefore water will be in liquid form.

(b) The water will be in gaseous form because from the figure we can see that the boiling temperature of water in pressure cooker is $120^circ text{C}$, and the given temperature is more than the boiling point therefore water will be a gas.

(c) The water will be in gaseous form because from the figure we can see that the boiling temperature of water on mountain top is $90^circ text{C}$, and the given temperature is more than the boiling point therefore water will be a gas.

(d) The water will be in liquid form because from the figure we can see that the boiling temperature of water on mountain top is $90^circ text{C}$, and the given temperature is less than the boiling point therefore water will be in liquid form.

Result

2 of 2

liquid, gas, gas, liquid

Result

1 of 1

(a) Liquid

(b) Gas

(d) Liquid

Exercise 53

Step 1

1 of 3

### Knowns

– The mass of the water $m = 0.96text{ kg}$

– The latent heat of fusion of water $L_f = 33.5 cdot 10^4 frac{text{J}}{text{kg}}$

– The initial and final temperature $T_i = T_f = 0text{textdegree}text{C}$

Step 2

2 of 3

### Calculation

To make ice cubes out of $m= 0.96text{ kg}$ of water we need to remove the following amount of thermal energy:

$$
begin{equation*}
Q = m , L_f
end{equation*}
$$

Plugging in the values we get:

$$
begin{align*}
Q &= 0.96text{ kg} cdot 33.5 cdot 10^4 frac{text{J}}{text{kg}} = 321600text{ J} \
Q &approx 3.2 cdot 10^5text{ J}
end{align*}
$$

Result

3 of 3

$$
begin{align*}
Q approx 3.2 cdot 10^5text{ J}
end{align*}
$$

Exercise 54

Step 1

1 of 3

### Knowns

– The mass of the water $m = 0.96text{ kg}$

– The latent heat of fusion of water $L-v = 22.6 cdot 10^5 frac{text{J}}{text{kg}}$

– The initial and final temperature $T_i = T_f = 100text{textdegree}text{C}$

Step 2

2 of 3

### Calculation

To make stream out of $m= 0.96text{ kg}$ water we need to add the following amount of thermal energy:

$$
begin{equation*}
Q = m , L_v
end{equation*}
$$

Plugging in the values we get:

$$
begin{align*}
Q &= 0.96text{ kg} cdot 22.6 cdot 10^5 frac{text{J}}{text{kg}} = 2169600text{ J} \
Q &approx 2.17 cdot 10^6text{ J}
end{align*}
$$

Result

3 of 3

$$
begin{align*}
Q approx 2.17 cdot 10^6text{ J}
end{align*}
$$

Exercise 55

Step 1

1 of 2

This energy is used to heat the ice from initial $T_i=-15^circ$ up to the melting point $T_0=0^circ$, then on melting the ice (latent heat will be denoted $lambda_m =334text{ kJ/kg}$ ) and finally on heating the water to required temperature ($T_f=15^circ$):

$$
Q=mc_i(T_0-T_i) + lambda_m m + mc_w(T_f-T_0)
$$

which yields

$$
m=frac{Q}{c_i(T_0-T_i) + lambda_m + c_w(T_f-T_0)}.
$$

Using $c_i =2text{ kJ/(Kg K)}$ and $c_w = 4.2text{ kJ/(Kg K)}$ we obtain

$$
m= 2.22text{ kg}.
$$

Result

2 of 2

Click here for the solution.

Exercise 56

Step 1

1 of 2

The thermal energy will be used only on melting the copper since it is already at its’ melting point:

$$
Q= lambda_m m =362.25text{ kJ}
$$

Result

2 of 2

Click here for the solution.

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