A pan filled with hot food cool from 94 -C to 86 -C in 2 minutes- When the root temperature is at 20 -C- the time taken to cool from 71 -C to 69 -C is-

The correct option is D 42 s From Newton's Law of cooling, #160; ln(T_1 T_0T_2 T_0)=kt Thus, ln(94 2086 20)=k(2× 60) And #160; ln(71 20 ...

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Standard XII

Physics

Conduction Law

A pan filled ...

Question

A pan filled with hot food cool from $94^{\circ} C$ to $86^{\circ} C$ in 2 minutes. When the root temperature is at $20^{\circ} C$ , the time taken to cool from $71^{\circ} C$ to $69^{\circ} C$ is:

50 s

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40 s

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38 s

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42 s

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Solution

The correct option is D $42 s$
From Newton's Law of cooling,
$l n (\frac{T_{1} - T_{0}}{T_{2} - T_{0}}) = k t$
Thus, $l n (\frac{94 - 20}{86 - 20}) = k (2 \times 60)$
And $l n (\frac{71 - 20}{69 - 20}) = k t^{'}$
$⟹ t^{'} = 42 s$

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A pan filled with hot food cool from 94∘C to 86∘C in 2 minutes. When the root temperature is at 20∘C, the time taken to cool from 71∘C to 69∘C is:

The correct option is D 42 sFrom Newton's Law of cooling, ln(T1−T0T2−T0)=ktThus, ln(94−2086−20)=k(2×60)And ln(71−2069−20)=kt′⟹t′=42 s

A pan filled with hot food cool from $94^{\circ} C$ to $86^{\circ} C$ in 2 minutes. When the root temperature is at $20^{\circ} C$ , the time taken to cool from $71^{\circ} C$ to $69^{\circ} C$ is:

The correct option is D $42 s$
From Newton's Law of cooling,
$l n (\frac{T_{1} - T_{0}}{T_{2} - T_{0}}) = k t$
Thus, $l n (\frac{94 - 20}{86 - 20}) = k (2 \times 60)$
And $l n (\frac{71 - 20}{69 - 20}) = k t^{'}$
$⟹ t^{'} = 42 s$