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Lab |
Vocabulary Test
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Why does ice in a glass of water make the water cold? |

As a substance changes from one phase to another, there will be a change in heat energy but NO temperature change. This happens because all the energy in the system is being used to change the phase and not to change the speed of particle motion.
Heat of Fusion:
- The amount of heat needed to change one gram of a substance from the solid phase to the liquid phase.
- The Hfus of water is about 80 calories / gram.
- The amount of heat needed to change one gram of a substance from the liquid phase to the gas phase.
- The Hvap of water is about 540 calories / gram.
Equations used to calculate heat:
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A typical problem involving change of phase and temperature:
1. First - Calculate the heat is needed to change the temperature of the 100 grams of ice from -25 oC to 0 oC using the equation
H = (mass) (temperature change) ( specific heat) 2. Second - With the temperature of the ice at 0 oC, calculate the heat needed to melt the ice using the equation
H = (mass) ( Hfus) 3. Third - You now have 100 grams of liquid water with a temperature of 0 oC. Calculate the heat needed to bring the temperature of the liquid water to 100 oC using the equation
H = (mass) (temperature change) ( specific heat) 4. Fourth - The temperature of the water is now 100 oC, calculate the heat needed to change its phase from liquid to gas using the equation
H = (mass) ( Hvap) 5. Finally - Add all the heat values together to answer the question.
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Homework Assignment 291:
This assignment must be turned in by the beginning of class tomorrow to receive credit.
Scoring criteria![]()
The "values" needed for these problems are given in the sample problem above.
- How much heat is needed to raise the temperature of 250 grams of water from -50 oC to -10 oC?
- How much heat is used to raise the temperature of 1 liter of liquid water by 40 Co?
Hint: Be sure you understand the wording of this problem.
- A sample of ice has a mass of 25 kilograms. Enough heat is added to the ice that it is eventually turned into steam. How much heat is used during just the phase changes that the sample goes through.
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When cooled, what common substance expands as it approaches its freezing point? |
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Insulation is a poor conductor of heat thereby reducing the transfer of heat that occurs by conduction and convection. The effectiveness of building insulating materials is measured according to R-value. |
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Ask your facilitator for a bimetallic strip and perform this experiment with your table.
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Test Your Concept Understanding: ![]()
- Examine how your home is heated then answer these questions:
- What type of heat system is used?
- Is more than one type used? If so, list them all.
- Make a floor plan of your house showing the location all the heating system parts.
- There are additional links at the bottom of the page to help with the following:
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What part of an internal-combustion engine turns linear motion into rotational motion? |

Heat engines are machines that burn some type of fuel to produce heat, then convert the heat energy into mechanical energy. Most heat engines burn gasoline
to produce heat. In addition to the fuel, large amounts of air
are needed for combusion.
Two types of heat engines:
1. External-Combustion Engine:
Fuel is burned outside the engine.
2. Internal-Combustion Engine:
Fuel is burned inside the engine.
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Any volunteers may ask the science facilitator for tools and remove the #1 piston from the Chevy 350 V-8 engine in the classroom. |
Research Links:
1. H = (mass) (Temp Change) ( Specific Heat)
2. H = (mass) ( Hfus)
3. H = (mass) (Temp Change) (Specific Heat)
4. H = (mass) ( Hvap)
A. H = (mass) (Temp Change) ( Specific Heat)
B. 1250 cal + 8000 cal + 10000 cal + 54000 + 2000 cal = 75250 calories
The temperature of a substance (ice in this case) does not change during a phase change because all the extra heat is used to change the phase. Ice absorbs heat from the water as it melts. This causes the water to get "cold" (have less heat).
H = (100 g) (25 Co) (0.5 cal/g . Co)
H = 1250 calories
H = (100 g) (80 cal/g)
H = 8000 calories
H = (100 g) (100 Co) (1 cal/g . Co)
H = 10000 calories
H = (100 g) (540 cal/g)
H = 54000 calories
You must manipulate the equation to yield Temperature Change
Temp Change = calories of heat / (mass) (Specific Heat)
Temp Change = 2000 calories / (100 g) (0.25 cal/g . Co)
Temp Change = 80 Co
Final Temperature = 100 oC + 80 Co
Final Temperature = 180 oC
75250 caloreis / 1000 calories per Kilocalorie = 75.25 KCAL