Question

Explain in detail Newton’s law of cooling.

Answer 1

Newton’s law of cooling: Newton’s law of cooling states that the rate of loss of heat of a body is directly proportional to the difference in the temperature between that body and its surroundings.

`"dQ"/"dt" ∝ ("T" - "T"_"s")` ............(1)

The negative sign indicates that the quantity of heat lost by liquid goes on decreasing with time. Where,

T = Temperature of the object

T_s = Temperature of the surrounding

       Cooling of hot water with time

From the graph in the figure, it is clear that the rate of cooling is high initially and decreases with falling temperature.

Let us consider an object of mass m and specific heat capacity s at temperature T. Let T_s be the temperature of the surroundings. If the temperature falls by a small amount dT in time dt, then the amount of heat lost is,

dQ = msdT .........(2)

Dividing both sides of equation (2) by dt

`"dQ"/"dt" = "msd"/"dt"` .........(3)

From Newton’s law of cooling `"dQ"/"dt" ∝ ("T" - "T"_"s")`

`"dQ"/"dt" = - "a"("T" - "T"_"s")` .............(4)

Where a is some positive constant. From equation (3) and (4)

− a (T − T_s) = `"ms" "dT"/"dt"`

`"dT"/("T" - "T"_"s") = -"a"/"ms" "dt"` .......(5)

Integrating equation (5) on both sides,

`int "dT"/("T" - "T"_"s") = - int "a"/"ms" "dt"`

ln (T − T_s) = `"a"/"ms" "t" + "b"_1`

Where b₁, is the constant of integration. Taking exponential both sides, we get

T = `"T"_"s" + "b"_2"e"^(-"a"/"ms""t")` ........(6)

here b₂ = `"e"^("b"_1)` = constant