An earthen pitcher loses 1 gm of water per minute due to evaporation- If the water equivalent of pitcher is 0-5 kg and pitcher contains 9-5 kg of water- Calculate the time required for the water in pitcher to cool to 28- C from original temperature of 30- C- Neglect radiation effects- Latent heat of vaporization in this range of temperature is 580 Cal - gm and specific heat of water is 1 Cal - gm - C- A- 30-5 minB- 41-2 minC- 38-6 minD- 34-5 min

The correct option is D 34.5 minQ=(ms Δ T)_W+(W Δ T)_P ∴ W= water equivalent =9.5 × 10^3 × (30 28) +0.5 × (10^3) (30 28) =(9.5+0.5)(2) × 10^3 Q_1=20 × 10^3 n ...

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Byju's Answer

Standard XII

Physics

Conduction Law

An earthen pi...

Question

An earthen pitcher loses $1 gm$ of water per minute due to evaporation. If the water equivalent of pitcher is $0.5 kg$ and pitcher contains $9.5 kg$ of water. Calculate the time required for the water in pitcher to cool to $28^{\circ} C$ from original temperature of $30^{\circ} C$ . Neglect radiation effects. Latent heat of vaporization in this range of temperature is $580 Cal/gm$ and specific heat of water is $1 {Cal/gm}^{\circ} C$ .

30.5 min

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41.2 min

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38.6 min

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34.5 min

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Solution

The correct option is D $34.5 min$
$Q = (m s Δ T)_{W} + (W Δ T)_{P}$
{ $∴ W =$ water equivalent}
$= 9.5 \times 10^{3} \times (30 - 28) + 0.5 \times (10^{3}) (30 - 28)$
$= (9.5 + 0.5) (2) \times 10^{3}$
$Q_{1} = 20 \times 10^{3}$ ... (1)
Heat lost by (water + pitcher) = Heat gained by water to evaporate.
$Q_{2} = m L = (\frac{d m}{d t} \times t) 580$
$\Rightarrow (\frac{1}{min}) t \times 580$ ... (2)
According to the given problem $Q_{2} = Q_{1}$
$580 t = 20 \times 10^{3}$
$t = \frac{20 \times 10^{3}}{580} = 34.5 min$

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