Chap. 8: Properties of Water

Question 1

What kind of interaction creates the surface tension in water?

Answer 1

Dipole-Dipole Interaction, which is also cohesion.

Question 2

Why does ice float?

Answer 2

Because the density of ice is less than the density of water.

• Density of ice < Density of water*
• D = m/v*

Question 3

Why is water so unique?

Answer 3

In the solid state, the particles of matter are usually much closer together than they are in the liquid state. So if you put a solid into its corresponding liquid, it sinks. But this is not true of water. Its solid state is less dense than its liquid state, so it floats.

Water’s boiling point is unusually high. Other compounds similar in weight to water have a muchlower boiling point.

Another unique property of water is its ability to dissolve a large variety of chemical substances. It dissolves salts and other ionic compounds, as well as polar covalent compounds such as alcohols and organic acids.

Water is sometimes called the universal solvent because it can dissolve so many things. It can also absorb a large amount of heat, which allows large bodies of water to help moderate the temperature on earth.

D solid > D liquid > D gas

Question 4

Exothermic

Answer 4

When a change (a chemical change, a physical change, etc.) releases heat energy.

Question 5

Endothermic

Answer 5

When a change (a chemical change, a physical change, etc.) absorbs heat or energy.

Question 6

Is ice melting an endothermic or an exothermic reaction?

Answer 6

Endothermic reaction, because it is absorbing heat.

Question 7

Is the combustion of your car exhaust an endothermic or exothermic reaction?

Answer 7

Exothermic reaction.

Your car is releasing gas and oxygen, which also releases energy.

Question 8

Units of Heat Energy

Answer 8

1. Calorie (cal) - The amount of energy required to raise the temperature of 1 gram of water by 1°C. (Heat calorie is written with a small c, while food calories are actually kilocalories, or Kcal, and are written with a capital C).
2. Joules (J) - 4.184 J = 1 cal for water.

Question 9

q = m • c • ΔT

Answer 9

The equation used in heat energy.

q = Heat energy in units of joules

m = Mass in units of grams

c = Specific heat in units of J/g°C

ΔT = Change in temperature (T_<em>f</em>–T_<em>i</em>)ºC

Question 10

44.0 g of aluminium absorbs 1870 J. The temperature increases from 25.0ºC to 72.5ºC. What is the specific heat?

Answer 10

q = m • c • ΔT

q = 1870 J (heat energy in units of joules)

m = 44.0 g (mass in units of grams)

c = ??? (specific heat energy in units of J/g°C)

ΔT = (72.5 – 25.0)ºC = 47.5ºC (change in temperature)

Since we are trying to find the specific heat, or c, we need to rearrange the original equation q = m • c • ΔT to c = q / m • ΔT and solve.

In conclusion: Water has a very high specific heat. Water absorbs large quantities of energy before getting “hot.”

Question 11

How do you find the specific heat of an unknown metal?

Answer 11

1. Find the mass using a scale.
2. Find the temperature using a thermometer.
3. Place the heated unknown metal into the styrofoam cup of water.
4. Wait a bit until the heat transfer is finished, i.e. when the heat from the unknown metal absorbs into the water and they are both the same temperatures (when T_f water = T_f metal).

• q metal (energy that has exited) = q water (energy that has been gained)*
• – q metal = + q water*
• exothermic = endothermic*

Question 12

20.0 g of iron at 225°C is placed into 51.0 g of water. The final temperature is 45.0°C. What was the initial temperature of the water?

Answer 12

This problem requires three different steps in order to find the initial temperature (T_i) of the water:

1. First, we must find the variables of the equation q = m • c • ΔT for both water and iron, and solve. To find the specific heat in unit of joules for water, or c, remember that it takes 4.184 J to heat up water 1°C. To find the specific heat in unit of joules for iron, look on Table 8.1 in the textbook (page 245).
2. Then, rearrange the equation q = m • c • ΔT to ΔT = q / m • c to get the change in temperature.
3. Then, use the equation ΔT = T_f – T_i to get the initial temperature of the water, or T_i. T_f is the final temperature of the water that was given, or 45.0°C.

Question 13

35.0 g of iron is heated from 22.0°C to 55.0°C, adding 513 J of heat. What is the specific heat in joules?

Answer 13

q = m • c • ΔT

Since we need to find the specific heat in joules, or c, we must rearrange the original equation into c = q / m • ΔT and solve. Remember that to get ΔT, you must subtract the final temperature from the initial temperature, or T_f – T_i , in which T_f = 55.0°C and T_i = 22.0°C.

Question 14

Calculate the energy in calories when 25.0 g of copper is heated from 23.0°C to 35.0°C.

The specific heat of copper is 0.385 J/g°c.

Answer 14

q = m • c • ΔT

Since we need to calculate the energy (in calories), we leave the equation as is and plug in the values to solve.

• q = (25.0 g) (0.385 J/g°C) (35.0–23.0)°C*
• q = 115.5 J*

We must now convert 115.5 J into calorie. Knowing that 1 calorie = 4.184 J, we can solve the rest of the equation.

Question 15

45.0 g of an unknown metal at 100°C is placed into 50.0 g of water at 25.0°C. The final temperature (or T_f) is 31.6°C. What is the unknown metal?

Answer 15

q = m • c • ΔT

This problem requires two different steps:

1. First, we must find the variables for the equation q = m • c • ΔT for both water and the unknown metal and figure out which one to solve first. Since water seems to have the most information, we solve for water first. To find the specific heat in unit of joules for water, or c, remember that it takes 4.184 J to heat up water 1°C.
2. Then, rearrange the equation q = m • c • ΔT to c = q /m • ΔT to get the specific heat in units of joules for the unknown metal. From there, you will find what the unknown metal is.

Question 16

When 47.5 J of heat is added to 13.2 g of a liquid, its temperature increases 17.2°C. What is the heat capacity of the liquid in J/g°C?

Answer 16

Since the problem is asking us for the heat capacity of the liquid in J/g°C, or c, we must rearrange the original equation q = m • c • ΔT to c = q / m • ΔT and solve.

Question 17

A 52.0 g sample of an unknown metal is heated to 125°C and placed into a container holding 65.0 g of water at 30.0°C. The metal and water reach a final temperature of 36.5°C.

– How much heat did the water absorb, in joules?

– What is the heat capacity of the metal in J/g°C?

– What is the likely identity of the unknown metal?

Answer 17

q = m • c • ΔT

First, we must find the variables of the equation q = m • c • ΔT for both water and iron, and solve. To find the specific heat in unit of joules for water, or c, remember that it takes 4.184 J to heat up water 1°C.

– How much heat did the water absorb, in joules?

It is asking for the amount of heat the water absorbed in joules, or q. We leave the original equation as is, and solve.

– What is the heat capacity of the metal in J/g°C?

It is asking for the heat capacity of the metal in J/g°C, or c, so we rearrange the original equation q = m • c • ΔT to c = q / m • ΔT and solve. Remember that in order to get the unknown metal’s heat energy, or q, it is the same as water’s heat energy, except it is a negative value (since the metal is losing heat, exothermic).

– Using the specific heat, what is the likely identity of the unknown metal?

We take the answer from the previous question to determine the unknown metal.