Question

Q23.State Newton’s law of cooling. Derive the expression.​

Answer 1

Answer:

For small temperature difference between a body and it's surrounding, the rate of the cooling of the body is directly proportional to the temperature difference and the surface area exposed. dQ/dt » ( q-qs), where q and qs are temperature corresponding to objects and surrounding.....

Explanation:

plz. mark me as brainliest and follow me.........

Answer 2

.______________________________________

$\huge{\underline{\mathrm\red{\fcolorbox{white}{white}{Answer}}}}$

''The rate of loss of heat of body is directly proportional to the difference of temperature of body and surroundings''

$\frac{ - dϙ}{dt} = Ꮶ(ᴛ _{2} - ᴛ _{1})$

Where K is positive constant depending upon the area and nature of surface of body to show that

.

$log_{e}( ᴛ_{2} - ᴛ_{1}) = -ʀ t + ᴄ$

using Newton's law of cooling.

Consider a body mass m and specific heat capacity is at temperature T2.Let the temperature of body falls by a small amount dT2.

The amount of heat lost by body is given by

dQ= msdT2➝ ( 1 )

Divide equation on both sides by dT,we get rate of loss of heat

.

$\frac{dϙ}{dt} = ms \: d \: \frac{ ᴛ_{2} }{dt} ➝(2)$

Let T1 be the temperature of surroundings From Newton's law of cooling,we have

$\frac{ dϙ}{dt} = - ᴋ(ᴛ _{2} - ᴛ_{1}) ➝(3)$

Using eq ( 2 ) and ( 3 )

$- ms \: d \: \frac{ᴛ _{2}}{dt} = ᴋ(ᴛ _{2} - ᴛ_{1}) ➝(4)$

By re arranging the terms

$\frac{d \: ᴛ _{2} }{ᴛ _{2} - ᴛ_{1} } = \frac{ -ᴋ }{ᴍ \: s} dt$

$\frac{d \: ᴛ _{2}}{ᴛ _{2} - ᴛ _{1} } = - ᴋ \: dt➝(5)$

By integrating eq ( 5 )on both sides

$∫\frac{d \:ᴛ _{2} }{ᴛ _{2} - ᴛ_{1}} = - ∫ᴋ \: dt➝(6) \\ \\ \frac{∫dᴛ}{ᴛ - a} = log_{e}(ᴛ - ᴄ)$

LHS of eq ( 6 )becomes

$log_{e}(ᴛ _{2} - ᴛ _{1} )$

using integration formula

$∫dt = ∫{t}^{0} dt \\ = \frac{ {f}^{0 + 1} }{0 + 1} \\ = t + c$

RHS of eq ( 6 ) becomes KT.

______________________________________

Eat 5 star do nothing