Suppose a cold beer at 40F is placed into a warm room at 70F. Suppose 10 minutes later,
the temperature of the beer if 48F. Use Newton's law of cooling to fi nd the temperature
25 minutes after the beer was placed into the room.
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Newton's Law of Cooling gives us that dT/dt = k*(T - Ta), where T is the temperature
at time t and Ta is the ambient room temperature. Separating this equation gives us
(1/(T - Ta)) dT = k dt. Now integrate both sides to find that
ln l T - Ta l = kt + C, and so l T - Ta l = e^(kt + C) = c*e^(kt), where I replaced
e^C by c. Now since in this case Ta > T we have l T - Ta l = Ta - T, and thus
Ta - T = c*e^(kt) -----> T = Ta - c*e^(kt). Now apply the given data.
At t = 0 we have T = 40, so since Ta = 70 we have that
40 = 70 - c*e^0 -----> c = 30. Next, at t = 10 we have
48 = 70 - 30*e^(10*k) -----> e^(10*k) = 22/30 = 11/15 ----> k = (1/10)*ln(11/15),
which makes k = -0.0310154 to 7 decimal places.
So T(25) = 70 - 30*e^(-0.0310154*25) = 56.184 degrees F to 3 decimal places.
that's a linear D.E. practice in this type. Y' + 8y = 32 p(x) = 8; q(x) = 32 I.F. = e^imperative of p(x)dx I.F. = e^8x subsequently: ye^8x = needed(32e^8xdx) + C ye^8x = 32/8 e^8x + C ye^8x = 4e^8x + C y = 4 + C e^(-8x); then y = 7 as x = 0 7 = 4 + Ce^0 C = 3 So, y = 4 +3e^(-8x)...end