## Music Video

The following humorous song from Mark Rosengarten describes the parts of the calorimetry formula and the experimental methods used to determine the joules of heat gained or lost by a change carried out in a measured mass of water.

What is the difference in temperature and heat?

## Answer:

Temperature is a measure of the average kinetic energy and does not depend upon the amount of matter in the sample. Heat is the total kinetic energy that flows because of a difference in temperature and does depend on the amount of matter.

Match the terms on the left to the items on the right. Click Answers under the table to check your responses.

1. kinetic | a. can be measured | |

2. potential | b. type of energy that is in motion | |

3. changes in energy | c. specific heat capacity | |

4. specific heat capacity | d. type of energy that is stored and cannot be measured | |

5. how much the temperature of substance rises depends on this | e. heat needed to raise the temperature of 1 gram of a substance by 1 degree C |

## Answers:

1. B , 2. D , 3. A , 4. E , 5. C

You can touch the aluminum pan of a TV dinner soon after is has been taken from the oven, but you will burn your hand if you touch the food it contains. Why is this so?

## Answer:

The aluminum has a lower specific heat than the food (specifically the water in the food) and it therefore heats up and cools off more quickly. A lot of heat must be released before the water will change its temperature even one degree.

Why doesn’t the temperature of water (for example) continually increase as it is heated?

## Answer:

The temperature will NOT increase during phase changes. During a phase change, the heat is making the solid turn to liquid or the liquid turn to steam rather than increasing the temperature.

What equations must be used to calculate the heat associated with a phase change?

## Answer:

Q = m x delta H_{vap} or Q = m x delta H_{fus}

Why can’t the specific heat equation be used?

## Answer:

Because there is no change in temperature.

Use these charts as needed in the following calculations:

Substance | Specific Heat |
---|---|

H_{2}O (l) |
4.184 |

H_{2}O (steam) |
2.02 |

Al (s) | 0.89 |

Fe (s) | 0.45 |

Water |
---|

ΔH_{fus} = 334 J/g |

ΔH_{vap} = 2260 J/g |

Click on the word *Answer* to see the solution to each problem.

How much heat is required to warm 275 g of water from 76 °C to 87 °C?

## Answer

Q = m x C x Δt

Δt = 87 °C = 76 °C = 11 °C

Q = 275 g x 4.184 J/g °C x 11 °C

Q = 1300 J

PCl_{3} is a compound used to manufacture pesticides. A reaction requires that 96.7 g of PCl_{3} be raised from 31.7 °C to 69.2 °C. How much energy will this require given that the specific heat of PCl_{3} is 0.874 J/g °C?

## Answer

Q = m x C x Δt

Δt = 69.2 °C – 31.7 °C = 37.5 °C

Q = 96.7 g x 0.874 J/g °C x 37.5 °C

Q = 3170 J

A quantity of water is heated from 25.0 °C to 36.4 °C by absorbing 325 J of heat energy. What is the mass of the water?

## Answer

Q = m x C x Δt

Δt = 36.4 °C – 25.0 °C = 11.4 °C

Q = 325 J

C = 4.184 J/g°C (at the temperature indicated, water is a liquid)

m = Q/ (C x Δt)

m = 325 J / (4.184 J/g°C(11.4°C))

m = 6.81 g

A 500. g sample of an unknown metal releases 6.4 x 10^{2} J as it cools from 55.0°C to 25.0°C. What is the specific heat of the sample?

## Answer

Q = m x C x Δt

Δt = 25.0°C – 55.0°C = 30.0°C

-6.4 x 10<sup>2</sup>J = (500g)(c)(-30.0°C)

C = (-6.4 x 10<sup>2</sup>J)/(500 g x -30.0°C)

C = 0.0427 J/g°C

This metal could now be identified through a table of known specific heat capacities.

In a household radiator, 1000. g of steam at 100.°C condenses (changes from gas to liquid). How much heat is released?

## Answer

Q = m x ΔH<sub>vap</sub>

Q = 1000. g x 2260 J/g

Q = 2,260,000 J

How much heat is necessary to change a 52.0 g sample of water at 33.0°C into steam at 110.0°C? This problem requires several steps since temperature changes and a phase change takes place. Use the hints to solve.

## Step 1

Solve for the heat required to increase the water temperature from 33.0°C to 100.0°C. Stop here because the water will change phase at this temperature.

Q = m x C x Δt

Q = 52.0 g x 4.184 J/g°C x 67°C

Q = 14577 J (Don’t round until the end)

## Step 2

Solve for the heat required to increase the water into steam (no change in temperature).

Q = m x ΔH<sub>vap</sub>

Q = 52.0 g x 2260 J/g

Q = 117520 J

## Step 3

Solve for the heat required to change the temperature of the steam from 100.0°C to 110.0°C.

Q = m x C x Δt

Q = 52.0 g x 2.02 J/g°C x 10°C

Q = 1050.4 J (Note a different C is used here because the chemical is steam, not the liquid water.)

## Step 4

To get the heat required for the whole process, add the calculated heats from above.

Q = 14577 J + 117520 J + 1050.4 J

Q = 133,000 J