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Specific Heat. What are the building blocks?. Specific Heat. This concept does not involve building or breaking down of molecules but might involve a mixture of elements or compounds. How water heats up or cools down is the standard for all other materials.

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### Winter Survival measuring the temperature rise of water. and a Safe Source of Heat for Field Operations

Specific Heat

This concept does not involve building or breaking down of molecules but might involve a mixture of elements or compounds.

How water heats up or cools down is the standard for all other materials.

There is usually an interest to how metals heat up. However, all materials heat up differently.

ENERGY

Energy measured in calories is used for this concept.

A calorie is defined asThe amount of heat energy needed to raise the temperature of one gram of water one degree Celsius.

Or the amount of heat energy lost when one gram of water cools one degree Celsius.

This calorie is different than the one you hear about for food. A food calorie is 1,000 times larger. It is often written with a capital “C”

Specific Heat involves a rate of heating based on weight

For water, there is one calorie for every one gram and for every degree Celsius.

One calorie for each gram and for each degree Celsius.

One calorie per gram per degree Celsius.

1 cal g oC

Different materials heat up at different rates.

90OC

90OC

1 kg of Aluminum

1 kg of Gold

20OC

20OC

Gold heats up about 7 times faster than Aluminum

Different materials store different amounts of heat energy.

90OC

1 kg of Water

90OC

1 kg of Gold

20OC

20OC

Water takes about 30 times longer to heat than gold, meaning it stores about 30 times more calories.

Explained by the difference of the specific heat of water versus soil.

Different materials heat up at different rates.

90OC

90OC

1 kg of Aluminum

1 kg of Gold

20OC

20OC

Gold heats up about 7 times faster than Aluminum

Different materials store different amounts of heat energy.

90OC

90OC

1 kg of Aluminum

1 kg of Gold

20OC

20OC

By the time aluminum heats up to 90OC it will have stored 7 times more calories of heat than the gold did.

Different materials store different amounts of heat energy.

90OC

1 kg of Water

90OC

1 kg of Aluminum

20OC

20OC

Water takes about 5 times longer to heat than aluminum, meaning it stores about 5 times more calories.

1 calorie is defined as the amount of energy needed to raise 1 gram of water 1 degree Celsius

90OC

1 kg of Water

1 kg of water would take 1,000 calories for each degree. Our 700C increase would take 1,000 x 70 = 70,000 calories.

20OC

1 mL or 1 cm 1 gram of water 1 degree Celsius3 (cc) of water has the mass of 1 gram.1,000 mL = 1 Liter has mass of 1 kilogram

1 kg of water would take 1,000 calories for each degree. Our 700C increase would take 1,000 x 70 = 70,000 calories.

90OC

1 kg of Water

1 Liter water

70,000 calories = 70 kilocalories

70,000 calories = 70 Calories

20OC

Food calories

The specific heat of water is 1 gram of water 1 degree Celsius1 calorie per 1 gram per 1 degree Celsius

1 cal g 0C

Specific Heat of Water

=

4.184 Joules = 1 calorie

Joule = The energy used by 1 watt for 1 second

__ g 1 gram of water 1 degree Celsius

1 cal g 0C

__ 0C

X

X

Let the dimensions tell you how to solve problems.

The specific heat of other materials can be calculated by measuring the temperature rise of water.

Blacksmiths may have been the first to recognize this approach.

Need modern blacksmith measuring the temperature rise of water.

The specific heat of other materials can be calculated by measuring the temperature rise of water.

- Grams
- Water
- Metal

- Initial and Final Temperatures
- Water
- Metal

Since specific heat is calories per gram per degree (raised/lowered) we need to keep track of what quantities?

Conservation of Energy measuring the temperature rise of water.

Energy Lost = Energy Gained

Calories Lost = Calories Gained

Joules Lost = Joules Gained

A request for investigation into the heat capacity of available materials

Problem with heating tents measuring the temperature rise of water.

- Heating the inside of tents is dangerous because of the potential for fire and carbon monoxide poisoning. It seems that heating objects outside the tent and then bringing them inside would be safer.

- Candidates considered for heating are metal objects like aluminum or iron.
- Rocks found in the vicinity could also be used.
- Water is another candidate.

Water: Pros and Cons aluminum or iron.

- On a gram for gram basis, water stores heat much more efficiently than any other candidates. However, water can only be brought to 100 degrees Celsius before it boils, and then it won’t get any hotter unless placed under pressure, which is dangerous.
- Water is also a precious commodity in some regions, so risking a spill may not be worth it.

Metal: Pros and Cons aluminum or iron.

- Some of the metal tools and equipment might be used. However, in a quick evacuation of camp it might cause problems if these items are too hot to handle.

Rocks: Pros and Cons aluminum or iron.

Rocks seem like the best candidate. However, we have no data on how efficient rocks can store heat, especially of rocks containing a mixture of minerals.

Heat energy released by rocks = Heat energy gained by tent. aluminum or iron.

There are stories of soldiers and others who said they heated up rocks, brought them into their tent, and the rocks warmed up the tent. This is good to know, but there is much information missing.

To calculate heat energy released from rocks we could calculate the heat energy gained by the tent, which is difficult because of the different materials in the tent. However, we could use waterinstead because we knowthat it takes one calorieto raise one gram of water one degree Celsius.

Heat energy released by rocks = Heat energy gained by water

The rock will begin to cool quickly after being removed from boiling water, so the transfer should be quick. Also shake off any excess water so heating is from rock and not from the hot water.

If heat gets transferred away from water, then the water won’t reach its full temperature. Therefore, insulation is critical. Use two Styrofoam cups. and a cover.

In practice, it wouldn’t be that easy to find a rock equal to one gram or to drop it into one gram of water. Therefore, the calculations will need to adjust for the mass of the rock and the mass of the water.

A rock could be heated by a flame but it would be difficult to measure the rock’s temperature. Another way would be to place the rock in boiling water and then measure the temperature of the water. Then the heated rock could be placed in a known amount of water and the temperature rise of the water could be recorded.

Should be near 100oC

Other heat sources could be used (hotplate, stove top, etc.)

Name: _________________ boiling water, so the transfer should be quick. Also shake off any excess water so heating is from rock and not from the hot water.

It’s February and you want to heat up your pool. It is currently 50oF (10oC) and you want it raised to 80oF (27oC) Using an electric heater, how much will it cost if 1 kilowatt-hour is 10 cents? The pool is 7.0 meters long, 5.0 meters wide and 2.0 meters deep.

A. What’s the volume in cubic meters? ______

B. What is the volume in cubic centimeters?___________

C. What is the total grams of the water? ____________

1 cc of water weighs 1 gram

A calorie is defined as the energy needed to raise one gram of water one degree Celsius. D. How many calories will it take to raise the pool water from 10oC to 27oC? _______________

E. 1 calorie = 1.1 x 10-6 kilowatts hours. How many kilowatt-hours will you need? ____________

F. How much will it cost to heat the pool from 50oF (10oC) to 80oF (27oC)? _______________

grams 17oC 1 calorie =

g. oC

calories 1.1x10-6 kilowatt-hrs =

_______ kilowatt-hours $0.10 =

calories

kilowatt-hr

grams x boiling water, so the transfer should be quick. Also shake off any excess water so heating is from rock and not from the hot water.1 cal x T g 0C

grams x SH x T =

Calories Lost = Calories Gained

calories lost by metal = calories gained by water

grams x 1 cal x (Tf - Ti) g 0C

grams x SH x (Tf - Ti) =

SH x g 0C amount = calories gained by water

g 0C amount g 0C amount

Specific Heatmetal = calories gained by metal

g 0C

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