A pork roast was taken out of a hardwood smoker when its internal temperature ha

Question

A pork roast was taken out of a hardwood smoker when its internal temperature had reached 190°F and it was allowed to rest in a 70°F house for 15 minutes after which its internal temperature had dropped to 175°F. Assume that the temperature of the roast follows Newton's Law of Cooling. Newton's Law of Cooling (Warming): The temperature T of an object at time t is given by the formula where T To is the initial temperature of the object, T, is the ambient temperature t and k > 0 is the constant of proportionality which satisfies the equation (instantaneous rate of change of T(t) at time t-k ()-T (a) Express the temperature T as a function of time t in minutes. ) = (b) Find the time at which the roast would have dropped to 140°F had it not been carved and eaten. (Round your answer to the nearest whole number.) min

Explanation / Answer

By Newton's law of cooling

S=70
at t=0

190 – 70 = C = 120
at t = 15

175 – 70 = 120*e^(-15k) = 105

e^(-15k) = 105/120
k = 0.008902

b) at T=140

140 – 70 = 70 = 120*e^(-kt)

70/120 = e^(-0.008902*t)

t = 60.5 minutes

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