Question

An earthen pitcher loses 1 gm of water per minute due to evaporation. If the water equivalent of the pitcher is 0.5 kg and the pitcher contains 9.5 kg of water, calculate the time required for the water in a pitcher to cool to 28°C from the original temperature of 30°C. Neglect radiation effects. The latent heat of vaporization in this range of temperature is 580 Cal/gm and the specific heat of water is 1 Cal/gm°C.

Options

• 30.5 min

• 41.2 min

• 38.6 min

• 34.5 min

Answer 1

34.5 min

Explanation:

Heat lost by (water + pitcher)

⇒ (9.5 + 0.5) × 10³ × (30 - 28)

⇒ 20 × 10³ cal

Heat gained for the water to evaporate:

(Let 't' be a time in minutes.)

⇒ (1 × t) × 580

⇒ 580 t Cal

580 t = `20 xx 10^3`

t = `20000/580` = 34.5 min