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 28⁰C from original temperature of 30⁰. Neglect radiation effects. Latent heat of vaporization in this range of temperature is 580 Cal/gm and specific heat of water is 1 Cal/gm⁰C.

• A. 30.5 min
• B. 41.2 min
• C. 38.6 min
• D. 34.5 min

Answer 1

Answer: (D)

34.5 min

Answer 2

Heat lost by (water + pitcher)

Ž (9.5 + 0.5) × 10³ × (30 – 28)

Ž 20 × 10³ cal

Heat gained for the water to evaporate :

(Let 't' be time in minutes)

Ž (1 × t) × 580

Ž 580 t Cal

580 t = 20 × 10³

t =$\frac{20000}{580}$= 34.5 min.