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The temperature of a pan of hot water varies according to Newton's Law of Cooling: dT dt equals negative k times the quantity T minus A , where T is the water temperature, A is the room temperature, and k is a positive constant. If the water cools from 90°C to 85°C in 1 minute at a room temperature of 30°C, how long (to the nearest minute) will it take the water to cool to 60°C?

User Johangu
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1 Answer

4 votes

Answer:7.96 min

Step-by-step explanation:

Given

time taken by water to cool from
90^(\circ)C to
85^(\circ)C is 1 min

Ambient Temperature
T_(\infty )=30^(\circ)C

According to Newtons law of cooling


(T-T_(\infty ))/(T_i-T_(\infty ))=e^(-kt)

where T=Final Temperature


T_i=Initial Temperature

k=time constant

t=time


(85-30)/(90-30)=e^2{-k\cdot 1}


(11)/(12)=e^(-k)


k=0.08701

Time taken to cool to
60^(\circ)C


(60-30)/(90-30)=e^(-kt)


(1)/(2)=e^(-kt)

taking Log


\ln 0.5=-kt


t=(0.6931)/(0.08701)


t=7.96 min

User Valloric
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