Advertisements
Advertisements
प्रश्न
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.
विकल्प
30.5 min
41.2 min
38.6 min
34.5 min
MCQ
रिक्त स्थान भरें
उत्तर
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
shaalaa.com
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?