Homework SourseA cup of coffee at 90 degrees C is put into a 20 degree C ro
A cup of coffee at 90 degrees C is put into a 20 degree C room when t=0.The coffee\'s temperature is changing at a rate of r(t)=-7e^(-0.1t) degrees C per minute, with t in minutes. Estimate the coffees temperature when t=10. Round answer to three decimal places.
Solution
Let T be the temperature (degrees Celsius) of the coffee, The rate of change of the temperature at time t was given by dT/dt = -7e^(-0.1t) Hence dT = -7e^(-0.1t) dt T = -7 ? e^(-0.1t) dt T = -7e^(-0.1t)(-0.1) + constant T = 0.7e^(-0.1t) + constant The constant is defined by the initial condition T(0) = 90 90 = 0.7e^(-0.1t) + constant constant = 90 - 0.7e^(-0.1x0)=89.3 Thus T = 0.7e^(-0.1t) + 89.3 At t = 10 T(10) = 0.7e^(-0.1x10) + 89.3 T(10) = 89.557 degrees C
