document.write( "Question 1189396: A car engine runs at a temperature of 87°C, when the engine is turned off, it cools according to Newton’s Law of Cooling (T=Ts+D0e-kt) with constant k=0.0347, where the time is measured in minutes. How long does it take for the engine to cool to 33°C if the surrounding temperature is 15°C. (A) 44min (B) 42min (C) 40min (D) 38min (E) 36min (note: ln2=0.6931)
\n" ); document.write( "How can i solve this without a calculator? \n" ); document.write( "
Algebra.Com's Answer #820746 by math_tutor2020(3817)\"\" \"About 
You can put this solution on YOUR website!

\n" ); document.write( "Given Info:
  • Ts = surrounding temperature (aka ambient temperature) = 15 degrees C
  • T0 = initial temperature of the object = 87 degrees C
  • k = 0.0347
  • t = time in minutes = unknown and what we want to solve for
  • D0 = T0 - Ts = difference in the initial temperature and the surrounding temperature
T = the temperature at time t
\n" ); document.write( "uppercase T for temperature, lowercase t for time\r
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\n" ); document.write( "\n" ); document.write( "The goal is to find lowercase t when uppercase T = 33\r
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\n" ); document.write( "\n" ); document.write( "T = Ts + D0e^(-kt)
\n" ); document.write( "T = Ts + (T0 - Ts)e^(-kt)
\n" ); document.write( "33 = 15 + (87 - 15)e^(-0.0347t)
\n" ); document.write( "33 = 15 + 72e^(-0.0347t)
\n" ); document.write( "33-15 = 72e^(-0.0347t)
\n" ); document.write( "18 = 72e^(-0.0347t)
\n" ); document.write( "e^(-0.0347t) = 18/72
\n" ); document.write( "e^(-0.0347t) = 1/4
\n" ); document.write( "Ln( e^(-0.0347t) ) = Ln(1/4)
\n" ); document.write( "-0.0347t*Ln(e) = Ln(2^(-2))
\n" ); document.write( "-0.0347t = -2*Ln(2)
\n" ); document.write( "-0.0347t = -2*0.6931
\n" ); document.write( "-0.0347t = -1.3862
\n" ); document.write( "t = -1.3862/(-0.0347)
\n" ); document.write( "t = 39.9481268011528
\n" ); document.write( "t = 40 minutes\r
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\n" ); document.write( "\n" ); document.write( "All of the calculations shown above are possible to do with pencil & paper only (i.e. without a calculator). It'll take a while, but it's still possible. \r
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\n" ); document.write( "\n" ); document.write( "Notice that I wrote 1/4 as 2^(-2), and then pulled down the exponent to get Ln(2) term. After which we substitute 0.6931 and evaluate. \r
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\n" ); document.write( "\n" ); document.write( "Answer: 40 minutes (choice C)
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