{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Lab 2 - The temperature conversion from degrees Fahrenheit...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Data: See attached results. Calculations: See attached results. Summary of Results and Conclusion: The attached table shows the decrease in temperature as it is recorded while decreasing two degrees Celsius for each recording. As shown in the table, the temperature decreased from 98.6 degrees Celcius to 68.0 degrees Celcius. It also decreased from 209 degrees Fahrenheit to 154.4 degrees Fahrenheit. We obtained a slope for our best fit line with both of our hand drawn graphs and a computer generated graph. The hand drawn graphs agree wih the computer generated graph in terms of predicting a temperature in degrees Fahrenheit from a given temperature in degrees Celsius. All three graphs predict that 165 degrees Fahrenheit is equivalent to roughly 74 degrees Celsius. The equation of the line for each graph is also remarkably similar.
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: The temperature conversion from degrees Fahrenheit to degrees Celsius is 5/9(F-32). The temperature conversion from degrees Celsius to degrees Fahrenheit is 9/5(C) + 32. The correct value for the slope in decimal form is 1.79 degrees Fahrenheit/degrees Celsius. Error Analysis: A possible source of error is a mulfunctioning temperature guage. This would cause our temperature results to be incorrect. Another source of error would be the level of heat that was used to boil the water. If the hottest part of the flame was not used, the water may not have heated to a maximum level thus changing the highest level at which the water would decrease from. When the temperature level decreased, it was not always recorded at exactly a two degree decrease. Sometimes the recorded decrease was less than two degrees but more than one....
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online