Heat of Fusion for Ice

Purpose:

Procedure:

1) Start with 100 ml H2O at fifty degrees Celsius

2) Place H2O into calorimeter and record temperature

3) Drop two to three ice cubes into calorimeter and stir until temperature no longer drops. Remove any remaining ice (don't run out of ice!)

4) Measure new volume

Data:

Temperature of H2O in calorimeter: 50 degrees Celsius

Volume of H2O after melting ice cubes: 163 ml

Calculations:

1) Mass of H2O used: 100 ml=100 g

2) Heat given by water: 100g x -48.7 x 4.18= -20,356.6 joules

3) Heat taken by ice: 20,356.6 joules

4) Mass of ice: 63g

5) Moles of Melted ice: 63g---> 3.5 moles

6) Heat Fusion: 20.4 kJ/3.5 mol ----> 5.8 kJ

Conclusion:

Practice Problems #13-15

Practice Problems #3-4

Practice Problems 27-20, 37-29

Practice Problems

Gay-Lussac's Law Lab

Graph:

Questions:

1) The constants were type of gas and volume of gas.

2) the relationship between pressure and temperature is that when temperature rises so does the pressure.

3) When water is heated, the molecules move around faster, leading to more collisions and production of energy, which excites the gas molecules in a similar way, causing both to move at greater speeds, increasing pressure.

4) K=P/T

5) Find constant how much did vary

6) It should also double if the temperature around it is doubled- The shown pressure values doubled in direct relation to the temperature.

Practice Problems #9-16