Finding the Heat of Vaporization of Liquid Nitrogen
Background:
The specific heat of water is the energy required to raise the temperature of 1 gram of water by 1 degree Celsius. Joules, grams, and celsius are used as units for this measurement because they are the standard units for measuring energy, mass, and temperature. The equation needed to calculate heat energy released by water in this experiment is q= energy (j) m= mass (g) T= temperature (C) c= specific heat capacity (q=mTc). The heat of energy lost by the hot water is the same energy gained by the liquid nitrogen. Heat of vaporization is the heat required to turn 1 gram of a substance from liquid to gas. The formula we use to calculate the heat of vaporization of a liquid is Total …show more content…
Place the styrofoam cup in a large (600 mL) beaker for stability. Find the mass of a second empty styrofoam cup. Your teacher will add about 80 mL of liquid nitrogen, N2 (l) to the cup. Find the mass of the nitrogen and the cup, then allow allow enough nitrogen to evaporate until a 60.0 g sample of nitrogen remains.Remember to take into account the mass of the cup - you want a 60.0 g sample of Nitrogen. Find temperature of the water to the nearest 0.l oC, then remove the thermometer. Quickly and carefully add the liquid nitrogen to the warm water. Observe. When the boiling stops, or when you no longer hear a "sizzling" sound, fan away any remaining …show more content…
The average mass of both hot and cold water is 90.6g. The change in temperature of the water is 33.1 degree Celsius. The total heat energy released by water was 12500J. The way we got 12500J was by (33.1 C) x (90.6 g) x 4.184 J/g C. Then next we found the mass of the liquid nitrogen at 60.0g. The heat gained per gram of nitrogen is 208 J/g. Dividing 12500J by 60g got us the answer of 208 J/g which the number we’re using. To find the molar heat of vaporization of nitrogen we multiplied 208 by 28.0 and got the answer of 5820 J/mol. The equation that we utilized to find the vaporization is N2(l) + 5820J. Finally, the equation for condensation is N2(l) + 5820J.Two errors in this experiment are heat lost to the air by the hot water, or heat gained by nitrogen for the cup and the air. These problems can be solved if we use better insulated. The percentage error we calculated was 4.3% we got the following anser by (5580-5820)/5580 x 100%.When nitrogen was added to the water water droplets were forming. The volume of the nitrogen gas was 47.5 L. The procedure we followed to the heat fusion of dry is was to add warm and cold water to two different Styrofoam cup and find the mass of both the liquids. Then calculate the change of temperature of the liquids, and then add dry ice to the warm water and measure the final temperature after it’s all gone. Finally to record the final temperature and the mass of
Background:
The specific heat of water is the energy required to raise the temperature of 1 gram of water by 1 degree Celsius. Joules, grams, and celsius are used as units for this measurement because they are the standard units for measuring energy, mass, and temperature. The equation needed to calculate heat energy released by water in this experiment is q= energy (j) m= mass (g) T= temperature (C) c= specific heat capacity (q=mTc). The heat of energy lost by the hot water is the same energy gained by the liquid nitrogen. Heat of vaporization is the heat required to turn 1 gram of a substance from liquid to gas. The formula we use to calculate the heat of vaporization of a liquid is Total …show more content…
Place the styrofoam cup in a large (600 mL) beaker for stability. Find the mass of a second empty styrofoam cup. Your teacher will add about 80 mL of liquid nitrogen, N2 (l) to the cup. Find the mass of the nitrogen and the cup, then allow allow enough nitrogen to evaporate until a 60.0 g sample of nitrogen remains.Remember to take into account the mass of the cup - you want a 60.0 g sample of Nitrogen. Find temperature of the water to the nearest 0.l oC, then remove the thermometer. Quickly and carefully add the liquid nitrogen to the warm water. Observe. When the boiling stops, or when you no longer hear a "sizzling" sound, fan away any remaining …show more content…
The average mass of both hot and cold water is 90.6g. The change in temperature of the water is 33.1 degree Celsius. The total heat energy released by water was 12500J. The way we got 12500J was by (33.1 C) x (90.6 g) x 4.184 J/g C. Then next we found the mass of the liquid nitrogen at 60.0g. The heat gained per gram of nitrogen is 208 J/g. Dividing 12500J by 60g got us the answer of 208 J/g which the number we’re using. To find the molar heat of vaporization of nitrogen we multiplied 208 by 28.0 and got the answer of 5820 J/mol. The equation that we utilized to find the vaporization is N2(l) + 5820J. Finally, the equation for condensation is N2(l) + 5820J.Two errors in this experiment are heat lost to the air by the hot water, or heat gained by nitrogen for the cup and the air. These problems can be solved if we use better insulated. The percentage error we calculated was 4.3% we got the following anser by (5580-5820)/5580 x 100%.When nitrogen was added to the water water droplets were forming. The volume of the nitrogen gas was 47.5 L. The procedure we followed to the heat fusion of dry is was to add warm and cold water to two different Styrofoam cup and find the mass of both the liquids. Then calculate the change of temperature of the liquids, and then add dry ice to the warm water and measure the final temperature after it’s all gone. Finally to record the final temperature and the mass of