Unit 17: Change of Phase

It takes 316 cal to change the ice cube to boiling water and 540 cal to change the boiling water to steam.

The amount of heat energy required to increase the temperature of 1 g of ice from absolute zero ($-{273}^{\circ }\text{C}$) to its melting point (0°C) is calculated as follows: $$\begin{array}{rl}{Q}_1& =(1\text{g})(0.5\text{cal}/\text{g}{\cdot }^{\circ }\text{C})({273}^{\circ }\text{C})\\ \text{(17.1)}& & =136\text{cal}\end{array}$$ The heat required just to melt the ice while remaining at 0°C is $$\begin{array}{rl}{Q}_2& =(1\text{g})(80\text{cal/g})\\ \text{(17.2)}& & =80\text{cal}\end{array}$$ To warm up 1 g of water from 0°C to 100°C, you require $$\begin{array}{rl}{Q}_3& =(1\text{g})(1\text{cal}/\text{g}{\cdot }^{\circ }\text{C})({100}^{\circ }\text{C})\\ \text{(17.3)}& & =100\text{cal}\end{array}$$ By adding the heat energies calculated above, you can determine the number of calories required to change a 1-g ice cube at absolute zero to boiling water: $$\begin{array}{rl}{Q}_{\text{total}}& =Q_1+Q_2+Q_3\\ & =136\text{cal}+80\text{cal}+100\text{cal}\\ \text{(17.4)}& & =316\text{cal}\end{array}$$ This is less than the amount of heat required to evaporate 1 g of water at 100°C: $$\begin{array}{rl}{Q}_{\text{vaporization}}& =(1\text{g})(540\text{cal/g})\\ \text{(17.5)}& & =540\text{cal}\end{array}$$ Note that it takes a lot of energy just to change the phase of water from the liquid state to the gaseous state.

Start by calculating the amount of heat energy lost by the poured water.

0°C.

Since the ice block is very large, the temperature of the water will drop until it reaches that of the ice (i.e., 0°C). The amount of heat energy ($Q$) lost by the water in the process is calculated as follows: $$\begin{array}{rl}Q& =(50\text{g})(1\text{cal}/\text{g}{\cdot }^{\circ }\text{C})({80}^{\circ }\text{C}-0^{\circ }\text{C})\\ \text{(17.6)}& & =4000\text{cal}\end{array}$$ This quantity of heat transfers to the surrounding ice and causes partial melting, with the temperature remaining at 0°C. Since each gram of ice requires 80 cal of latent heat to melt, you can calculate the mass ($m$) of melted ice as follows: $$\begin{array}{rl}m& =\frac{4000\text{cal}}{80\text{cal/g}}\\ \text{(17.7)}& & =50\text{g}\end{array}$$

The amount of heat gained by 2 kg of ethyl alcohol when it evaporates is calculated to be $$\begin{array}{rl}Q& =(2000\text{g})(200\text{cal/g})\\ \text{(17.8)}& & =\mathrm{400,000}\text{cal}\end{array}$$ Assuming all this heat is taken from a certain quantity of water at 0°C, causing it to freeze, the mass ($m$) of the ice formed is calculated to be $$\begin{array}{rl}m& =\frac{Q}{80\text{cal/g}}\\ & =\frac{\mathrm{400,000}\text{cal}}{80\text{cal/g}}\\ \text{(17.9)}& & =5000\text{g}(\text{or}5\text{kg})\end{array}$$

By sweating.

Human skin contains numerous sweat glands. High temperatures provoke sweating, and the sweat produced evaporates by drawing heat from the body, causing the body to cool.

Because of the condensation effect.

Air in the vicinity of an iceberg is chilled by the iceberg’s freezing temperature and becomes cooler than the surrounding air above the ocean. As a result, a cold spot is formed above the iceberg, whose relative humidity increases to the level of saturation. Any excess water vapor then condenses and appears as fog.

It releases thermal energy.

One gram of boiling water requires 540 cal of heat to evaporate and change into steam. This is referred to as the latent heat of vaporization. The same quantity of heat is released when 1 g of water vapor changes phase and becomes liquid. Therefore, condensation in the clouds releases a large amount of thermal energy.

Because of the reduction in liquid pressure.

The pressure of the liquid on an air bubble decreases as the bubble rises in the pot. As a result, the air inside the bubble expands.

Think about the relations between pressure, the boiling temperature of water, and cooking time.

No. Changes that occur to food during the cooking process are caused by increased temperature over a specific period of time. For example, at sea level, it takes about 10 minutes at a temperature of 100°C to produce a hard-boiled egg. So if you live in Vancouver, this is how much time you will need to prepare your breakfast. However, if you take the Skyride to the top of Grouse Mountain (1200 m above sea level), you will notice that water boils at 96°C. The reduced temperature of the boiling water would make the egg take a slightly longer time to cook.

The problem becomes more severe at the top of Mt. Logan (the highest summit in Canada). At an altitude of nearly 6000 m, water boils at 79°C. An egg would take about 25 minutes to cook at this elevation (see this interesting egg boiling calculator). Note that at the top of Mt. Everest (altitude of 8850 m), water boils at 68°C, which may not be a sufficient temperature for cooking eggs!

So as the atmospheric pressure drops with the increased elevation above sea level, water boils at lower temperatures. In a vacuum, the pressure is reduced to zero, so water boils even at room temperature. Therefore, the boiling process by itself is irrelevant because it does not transfer heat to the food. On the surface of the Moon, where there is no atmosphere, water boils spontaneously without heating, making it ineffective for cooking an egg.

Because of the sudden drop in pressure.

The water inside a closed radiator expands as the engine heats up, which causes the pressure to increase and become greater than atmospheric pressure. As a result, the temperature can exceed 100°C without causing the water to boil. When the radiator cap is removed, the pressure inside the radiator suddenly drops to normal atmospheric pressure, which causes water to boil vigorously, sometimes even explosively.

Yes.

When you place a pail of water outside your home on a cold winter night, the water cools down gradually until the temperature reaches 0°C, when ice starts to form. At this temperature, ice and water can coexist as a mixture. After all the water freezes, the ice temperature continues to fall below 0°C.

On February 3, 1947, the temperature in Snag, Yukon, set the record for the coldest temperature ever recorded in Canada, at $-{63}^{\circ }\text{C}$. A piece of ice in such an environment would reach the same temperature.

Because of the large amount of latent heat released by freezing water.

When 1 g of water freezes, 80 cal of latent heat are released into the surrounding environment. As a result, the temperature inside the cellar will not fall below zero as long as the water in the tub is not completely frozen. A 200-L bathtub (or 200,000 g of water) will release 16,000,000 cal of heat before it is completely frozen.

Melting ice decreases the temperature of the surrounding air, and freezing water increases it.

When ice melts, it changes phase from solid to liquid, which requires a significant amount of thermal energy (heat). In the absence of a direct heat source (such as the Sun or a stove), the required thermal energy is absorbed from the surrounding environment, including the air, which causes the air temperature to drop.

The opposite is also true. During the freezing process, a significant amount of heat is released into the surrounding environment (see the answer and explanation for Question 98).

Consider how the ability of air to hold water vapor changes with temperature.

This is due to condensation of water vapor. Warm, humid air contains a fair amount of water vapor. However, the ability of air to hold moisture decreases at lower temperatures. As it comes in contact with the cold components of an air-conditioning unit, warm air cools down and reaches saturation, at which point excess water vapor condenses, then drips down the unit. This is similar to the formation of water droplets on a glass of ice water on a hot, humid day.