The temperature M(t) outside a building decreases at a constant rate of 1degree

Question

The temperature M(t) outside a building decreases at a constant rate of 1degree Celcius per hour. The inside of the building is heated, and there is no other source of cooling. The heater was switched on at time t=0, when the temperature inside, T(t), was 17 degrees Celcius, and the temperature outside was 0 C. Assume that the heater generates a constant amount h= 50 000 Btu/hr of heat when it is working, the heat capacity of the building is r=1/5 degrees per thousand Btu and the time constant for heat transfer between the outside and inside of building is g=2 hr. On the basis of Newton's law of cooling,

dT(t)/dt = K(M(t) - T(t)) + rh,

Find the upper value of the temperature in the building in the time interval 0 <= t < 4 hr

Explanation / Answer

K=1/time constant = 1/2 = 0.5

Now,as the outside temp. is decreasing at a const. rate of 1 degree celcius per hour

M(t) = 0-1.t = -t (where t is in hour)

T(t) = 17+r.h.t= 17+10t

Now putting K,M(t) and T(t) in the Newton's law of cooling eqn.

dT(t)/d(t) = 0.5(-17-11t) + 10

dT(t)/d(t) = 1.5-6.5t

Now the temp in the building will be max when

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