A liquid takes 5 minutes to cool from $80^{0} C t o 50^{0} C$ . How much time will it take to cool from $60^{0} C t o 30^{0} C$ ? The temperature of the surrounding is $20^{0}$ C .

5 minutes

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9 minutes

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4 minutes

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12 minutes

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Solution

The correct option is B
9 minutes

According to Newton’s law of cooling, $(\frac{θ_{1} - θ_{2}}{t}) = K [(\frac{θ_{1} + θ_{2}}{2} - θ_{0})] S o t h a t (\frac{80 - 50}{5}) = K [(\frac{80 + 50}{2} - 20)] A n d (\frac{60 - 30}{t}) = K [(\frac{60 + 30}{2}) - 20]$

Solving these for t, we get t=9 minute.

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Q. A liquid takes 10 minutes to cool from

80^{o} C

50^{o} C

Calculate the time it takes to cool from

60^{0} C

30^{0} C .

The temperature of the surroundings is

20^{0} C .

If a body takes 6 minutes to cool from 80⁰C to 70⁰C, how long will it take to cool from 60⁰C to 50⁰C, if the surrounding temperature is 25⁰C?

A liquid takes 5 minutes to cool from $80^{0} C t o 50^{0} C$ . How much time will it take to cool from $60^{0} C t o 30^{0} C$ ? The temperature of the surrounding is $20^{0}$ C .

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A liquid takes 5 minutes to cool from 800 C to 500 C . How much time will it take to cool from 600 C to 300 C ? The temperature of the surrounding is 200 C .

The correct option is B 9 minutes According to Newton’s law of cooling, (θ1−θ2t)=K[(θ1+θ22−θ0)]So that(80−505)=K[(80+502−20)]And (60−30t)=K[(60+302)−20] Solving these for t, we get t=9 minute.

A liquid takes 5 minutes to cool from $80^{0} C t o 50^{0} C$ . How much time will it take to cool from $60^{0} C t o 30^{0} C$ ? The temperature of the surrounding is $20^{0}$ C .