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.

(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.

### 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 $T_i = T_f = 0text{textdegree}text{C}$

### 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:

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

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

### 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 $T_i = T_f = 100text{textdegree}text{C}$

### 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:

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

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

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

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

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

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