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एक पिण्ड 7 मिनट में `60^(@)C` से `40^(@)C` तक ठण्डा होता है। परिवेश का ताप `10^(@)C` है। अगले 7 मिनट पश्चात इसका ताप क्या होगा? पूरी प्रक्रिया में न्यूटन का शीतलन नियम लागू रहता है।

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माना किसी क्षण पिण्ड का ताप T तथा परिवेश का ताप `T_0` है तब पिण्ड से ऊष्मा हानि की दर
`" " -mc (dT)/(dt) =k(T-T_0) `
ताप गिरने की दर `" " -(dT)/(dt)=K(T-T_0)`
यहाँ `K=(k)/(mc) ` तथा ` T=(T_1+T_2)/(2) `
`" " therefore -(60-40)/(7xx60) =K ((60+40)/(2)-10) `
अथवा `" " -(20)/(420) =K xx40`
दूसरी अवस्था में ` T_1=40^(@)C,T_2=? `
`" " therefore -(40-T_2)/(420) =K ((40+T_2)/(2)-10)" "...(ii) `
भाग करने पर
`" " (20)/(40-T_2) =(40)/((40+T_2)/(2)-10)`
हल करने पर
`" " T_2 =28^(@) C.`

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