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Initial temperature $(T_{i})=80_{∘}$C

Final temperature $(T_{f})=50_{∘}$C

Temperature of the surrounding $(T_{0})=20_{∘}$C

$t=5$ min

According to Newton's law of cooling,

Rate of cooling $dtdT =K[2(T_{i}+T_{f}) −T_{o}]$

$t(T_{f}−T_{i}) =K[2(80+50) −20]$

$580−50 =K[65−20]$

$6=K×45$

$K=456 =152 $

In second condition,

initial temperature $=T_{i}=60_{∘}$C

Final temperature $T_{f}=30_{∘}$C

Time taken for cooling is $t$

According to Newton's law of cooling

$t(60−30) =152 [2(60+30) −20]$

$t30 =152 ×25$

$t30 =1550 =310 $

$∴t=9$ min

Solve any question of Thermal Properties Of Matter with:-

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