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System, Surroundings, Intensive, Extensive Properties

Two bottles A...

Question

Two bottles A and B have radii $R_{A}$ and $R_{B}$ and heights $h_{A}$ and $h_{B}$ respectively with $R_{B} = 2 R_{A}$ and $h_{B} = 2 h_{A}$ . These are filled with hot water at $60^{o}$ C. Consider that heat loss for the bottles takes place only from side surfaces. If the time the water takes to cool down to $50^{o}$ C is $t_{A}$ and $t_{B}$ for bottles A and B, respectively, then $t_{A}$ and $t_{B}$ are best related as.

A

t_{A} = t_{B}

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B

t_{B} = 2 t_{A}

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C

t_{B} = 4 t_{A}

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D

t_{B} = t_{A} / 2

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Solution

The correct option is A $t_{B} = 2 t_{A}$
From Newton's Law of cooling,
$\frac{d Q}{d t} = h A Δ T$

Assuming, constant rate of cooling,
$Δ Q = h A Δ T Δ t$
$\frac{Δ Q_{A}}{Δ Q_{B}} = \frac{A_{A} t_{A}}{A_{B} t_{B}}$
$\frac{m_{A} s Δ T}{m_{B} s Δ T} = \frac{A_{A} t_{A}}{A_{B} t_{B}}$
$\frac{ρ V_{A}}{ρ V_{B}} = \frac{A_{A} t_{A}}{A_{B} t_{B}}$
$\frac{h_{A} A_{A}}{h_{B} A_{B}} = \frac{A_{A} t_{A}}{A_{B} t_{B}}$
$\frac{t_{A}}{t_{B}} = \frac{1}{2}$
$t_{B} = 2 t_{A}$

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