a. The normal body temperature for cats varies over a narrow range of temperatur

Question

a. The normal body temperature for cats varies over a narrow range of temperatures. One cat has a history of a body temperature of 39.1 C . One night this cat is hit by a car at some time during the night. The dead cat is discovered at 7 AM, and a young scientist wants to determine the time of death of the poor cat. He measures the temperature of the cat and finds that the body temperature of the cat is 22.8 C (H(0)=22.8). For this problem, t=0 corresponds to 7 AM. The early morning temperature is found to be 15.5 C. Newton's Law of Cooling with a constant environmental temperature (Te=15.5) gives the differential equation: dH/dt= -k1(H-15.5), H(0)= 22.8 where H(t) is the body temperature of the cat and k1 is a kinetic constant of cooling. Find the solution to this differential equation with its initial condition (including the cooling constant k1 written as 'k1'). H(t)= _______________ If one hour later the temperature of the body is found to be 20.7 C (H(1)= 20.7), then determine the value of the constant of cooling, k1, in the differential equation. k1= _______________ Find the time "td" when the death occurs and give the time on the clock. td= _______________ . Time on the clock = ________ : __________ AM, where the minutes are a decimal value with at least 4 significant figures.

Explanation / Answer

dH/dt = -k1*(H - 15.5)

dH / (H - 15.5) = -k1*dt

Integrating both the sides,

ln(H - 15.5) = -k1*t + C

Now at t = 0, we have H(0) = 22.8

Thus, ln(22.8 - 15.5) = -k1*0 + C

C = 1.9878

Therefore,

ln (H - 15.5) = -k1*t + 1.9878

(H - 15.5) = e^[1.9878 - k1*t]

H(t) = 15.5 + e^(1.9878 - k1*t).........1

b)

Now H(1) = 20.7

Putting t = 1 in eqn 1

20.7 = 15.5 + e^(1.9878 - k1*1)

Solving this, k1 = 0.339

c)

For H(t) = 39.1 we get

39.1 = 15.5 + e^(1.9878 - 0.339*t)

Solving this, t = -3.461275 hours = -3 hours 27.6765 mins

Since t = 0 is at 7 AM, we get clock time = 7 - 3.461275 = 3 hours 32.3235 mins AM

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