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# A Hot Body Placed in a Surrounding of Temperature θ0 Obeys Newton'S Law of Cooling D θ D T = − K ( θ − θ 0 ) . Its Temperature at T = 0 is θ1. - Physics

#### Question

A hot body placed in a surrounding of temperature θ0 obeys Newton's law of cooling (d theta)/(dt) = -K(theta - theta_0)  . Its temperature at t = 0 is θ1. The specific heat capacity of the body is sand its mass is m. Find (a) the maximum heat that the body can lose and (b) the time starting from t = 0 in which it will lose 90% of this maximum heat.

#### Solution

According to Newton's law of cooling,

(d theta)/(dt) = -K (theta - theta_0)

(a) Maximum heat that the body can lose, ΔQmax = ms (θ1 - θ0)

(b) If the body loses 90% of the maximum heat, then the fall in temperature will be θ

ΔQ_maxxx90/100 = ms (theta_1 - theta)

⇒ ms (theta_1 - theta_0)xx9/10 = ms (theta_1 - theta)

⇒ θ = θ1 - (θ10) × 9/10

⇒ θ = (theta_1 - 9theta_0)/10..............(i)

From Newton's law of cooling,

(d theta)/(dt) = -K(theta_1 - theta)

Integrating this equation within the proper limit, we get

At time t = 0,

θ = θ1

At time t,

θ = θ

int_ {θ 1}^θ  (dθ)/(θ _1 - θ ) = -K int_0^t dt

rArr In (theta_1 - theta)/(theta_1 - theta_0) = -kt

⇒ θ_1 - θ= θ_1 - θ_0e^-"kt"...........(ii)

From (i) and (ii),

(θ_1 - 9θ_0)/10 - θ_0 = (θ_1 - θ_0)e^-kt

⇒  t =(In (10))/k

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Solution A Hot Body Placed in a Surrounding of Temperature θ0 Obeys Newton'S Law of Cooling D θ D T = − K ( θ − θ 0 ) . Its Temperature at T = 0 is θ1. Concept: Qualitative Ideas of Blackbody Radiation.
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