o \\%\\ [ N S VL S ८. 5 15 omiodaN D) Qonen\axe R« MoL 60" ४ 220

1.	o \\%\\ [ N S VL S ८. 5 15 omiodaN D) Qonen\axe R« MoL 60" ४ 220 ८. Ve vecaoeeds Bata ik vt N SR
Answer» Here, Initial temperature ( Ti) = 80°C Final temperature ( Tf) = 50°C Temperature of the surrounding ( To) = 20°C t = 5 min A/C to Newton's law of cooling Rate of cooling ( dT/dt) = K[ (Ti+Tf)/2 - To] ( Tf - Ti)/t = K[ ( 80 + 50)/2 - 20]( 80-50)/5 = K[ 65 - 20]6 = K× 45 K = 6/45 = 2/15 in second condition, initial temperature ( Ti) = 60°C Final temperature ( Tf) = 30°C Time taken for cooling is t A/C Newton's law of cooling ( 60 - 30)/t = 2/15 [ (60+30)/2 -20]30/t = 2/15 × 25 30/t = 50/15 = 10/3 t = 9 min

o \\%\\ [ N S VL S ८. 5 15 omiodaN D) Qonen\axe R« MoL 60" ४ 220 ८. Ve vecaoeeds Bata ik vt N SR

Answer»

Here, Initial temperature ( Ti) = 80°C Final temperature ( Tf) = 50°C Temperature of the surrounding ( To) = 20°C t = 5 min A/C to Newton's law of cooling

Rate of cooling ( dT/dt) = K[ (Ti+Tf)/2 - To] ( Tf - Ti)/t = K[ ( 80 + 50)/2 - 20]( 80-50)/5 = K[ 65 - 20]6 = K× 45 K = 6/45 = 2/15

in second condition, initial temperature ( Ti) = 60°C Final temperature ( Tf) = 30°C Time taken for cooling is t A/C Newton's law of cooling ( 60 - 30)/t = 2/15 [ (60+30)/2 -20]30/t = 2/15 × 25 30/t = 50/15 = 10/3 t = 9 min

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