document.write( "Question 1078775: a dish of lasagna baked at 350 degrees is taken out of the oven into a kitchen that is 68 degrees. After 7 minutes, the temperature of the dish is 299.8 degrees. what will its temperature be 12 minutes after it was taken out of the oven? Round to the nearest degree. \n" ); document.write( "
Algebra.Com's Answer #693224 by jorel1380(3719) You can put this solution on YOUR website! Newton’s Law of Cooling Formula is expressed by\r \n" ); document.write( "\n" ); document.write( "T(t) = Ts + (T0 – Ts) e-Kt\r \n" ); document.write( "\n" ); document.write( "Where,\r \n" ); document.write( "\n" ); document.write( "t = time,\r \n" ); document.write( "\n" ); document.write( "T(t) = temperature of the given body at time t,\r \n" ); document.write( "\n" ); document.write( "Ts = surrounding temperature,\r \n" ); document.write( "\n" ); document.write( "To = initial temperature of the body,\r \n" ); document.write( "\n" ); document.write( "k = constant. \n" ); document.write( " \n" ); document.write( "So: \n" ); document.write( "T(7)=68+(350-68)e^-kt \n" ); document.write( "299.8=68+282e^-7k \n" ); document.write( "e^-7k=0.82198581560283687943262411347518 \n" ); document.write( "ln e^-7k=ln 0.82198581560283687943262411347518 \n" ); document.write( "-7k=-0.1960321400324622534737708260656 \n" ); document.write( "k=0.02800459143320889335339583229509\r \n" ); document.write( "\n" ); document.write( "Then: \n" ); document.write( "T(12)=68+(350-68)e^-12(0.02800459143320889335339583229509) \n" ); document.write( "T(12)=68+(282)0.71458373316960990812286116751929 \n" ); document.write( "T(12)=269.512 degrees 12 minutes out of the oven. ☺☺☺☺\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |