Question #104078
Newton’s law of cooling assumes that air at room temperature is blown past the cooling body (forced cooling). For cooling in still air (natural cooling) a better modal is to assume that the rate of temperature decrease of the cooling body is directly proportional to the the (5/4)th power of the difference between the temperature u of the body and the temperature s of the surrounding air.
i) Write the law for natural cooling as a differential equation. Is this equation linear?
ii) Solve the equation obtained in i) above assuming that initially, the temperature of the cooling
body was u0.
1
Expert's answer
2020-03-10T11:19:49-0400

du/dt=k(us)5/4-du/dt=k(u-s)^{5/4}

is the differential equation for law of natural cooling.

Now, as degree of the differential equation is 1, thus it is linear.

u0udu/(us)5/4=0tkdt\int_{u_0}^u du/(u-s)^{5/4}=-\int_0^t k dt

(us)1/4(u0s)1/4=4kt(u-s)^{-1/4} -(u_0 -s)^{-1/4}=-4kt

u=s+((u0s)1/44kt)4u=s+((u_0 -s)^{-1/4}-4kt)^{-4}

is the solution for the above differential equation.


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Comments

Assignment Expert
08.03.20, 20:07

Dear Raj, please use the panel for submitting new questions.

Raj
08.03.20, 04:15

According to Newton’s law of cooling, the rate at which a substance cools in moving air is proportional to the difference between the temperature of the substance and that of the air. If the temperature of the air is C o 290 and the substance cools from C o 370 to C o 330 in 10 minutes, find when the temperature will be C o 295

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