A hot ball cools from 90 °C to 10 °C in 5 minutes. If the surrounding temperature is 20°C, what is the time taken to cool from 60 °C to 30 °C?
Here, Initial temperature (Ti)=80∘C Final temperature (Tf)=50∘C Temperature of the surrounding (T0)=20∘C t=5 min According to Newton's law of cooling, Rate of cooling dtdT=K[2(Ti+Tf)−To] t(Tf−Ti)=K[2(80+50)−20] 580−50=K[65−20] 6=K×45 K=456=152 In second condition, initial temperature =Ti=60∘C Final temperature Tf=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
