Question

# A body cools from $${80}^{\circ}$$C to $${50}^{\circ}$$C in $$5$$ minutes. Calculate the time it takes to cool from $${60}^{\circ}$$C to $${30}^{\circ}$$C. The temperature of the surroundings is $${20}^{\circ}$$C.

Solution

## Here, Initial temperature $$( {T}_{i}) = {80}^{\circ}$$C Final temperature $$( {T}_{f}) = {50}^{\circ}$$C Temperature of the surrounding $$( {T}_{0}) = {20}^{\circ}$$C $$t = 5$$ min According to Newton's law of cooling,Rate of cooling $$\dfrac{dT}{dt} = K[\dfrac{\left({T}_{i}+{T}_{f}\right)}{2}-{T}_{o}]$$$$\dfrac{\left({T}_{f}-{T}_{i}\right)}{t}=K\left[\dfrac{\left(80 +50\right)}{2} - 20\right]$$$$\dfrac{80-50}{5} = K[ 65 - 20]$$$$6 = K\times 45$$$$K = \dfrac{6}{45} = \dfrac{2}{15}$$ In second condition, initial temperature $$={T}_{i}= {60}^{\circ}$$C Final temperature $${T}_{f}={30}^{\circ}$$C Time taken for cooling is $$t$$ According to Newton's law of cooling $$\dfrac{( 60 - 30)}{t} = \dfrac{2}{15} \left[ \dfrac{\left(60+30\right)}{2} -20\right]$$$$\dfrac{30}{t}=\dfrac{2}{15}\times 25$$$$\dfrac{30}{t} = \dfrac{50}{15}=\dfrac{10}{3}$$$$\therefore t = 9$$ minPhysics

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