document.write( "Question 1129390: Can someone please help explain how to solve the problem:
\n" ); document.write( "A pot of boiling soup with an internal temperature of 100° Fahrenheit was taken off the stove to cool in a 69°F room. After fifteen minutes, the internal temperature of the soup was 93°F. \r
\n" ); document.write( "\n" ); document.write( "To the nearest minute, how long will it take the soup to cool to 81°F?
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Algebra.Com's Answer #745957 by solver91311(24713)\"\" \"About 
You can put this solution on YOUR website!
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\n" ); document.write( "\n" ); document.write( "Newton's Formula for Cooling:\r
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\n" ); document.write( "\n" ); document.write( "where is the temperature of the object at time , is the temperature of the surrounding environment, is the initial temperature of the object, and is a constant related to the nature of the object.\r
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\n" ); document.write( "\n" ); document.write( "First, we need the value of for this particular flavor of soup.\r
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\n" ); document.write( "\n" ); document.write( "Solve for , then substitute that value into\r
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\n" ); document.write( "\n" ); document.write( "and then solve for which will be the elapsed time in minutes since cooling began. Subtract 15 if you want the time between when the soup was 93 degrees and when it cooled to 81 degrees.\r
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\n" ); document.write( "\n" ); document.write( "The math works out with this one, but there is one huge problem with the way this question is posed as it relates to the real world. Soup (or any other water-based liquid for that matter) does not boil at 100 degrees Fahrenheit. It boils at 212 degrees Fahrenheit or 100 degrees Celsius. 93 degree F soup would be pee-warm, and 81 degree F soup would be cold. Remember, your body is 98.6 degrees F if you are reasonably healthy. So whoever wrote this question is suffering from a severe case of recto-cranial inversion.
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\n" ); document.write( "John
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\n" ); document.write( "My calculator said it, I believe it, that settles it
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