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One gram of water of volume 1 c.c becomes 1671 c.c of steam when boiled at a pressure of one atmosphere. The latent heat of vaporisation of water is 540 cal/gr. Find the increase in internal energy (1 at. pressure = 10⁵ N/m²)

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540 Cal

500.24 Cal

167 Cal

39.66 Cal

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detailed solution

Correct option is B

ΔU=Q−WHere W=P×V2−V1=105×(1671−1)×10−8J=167Jand Q = 1x540cal∴ΔU=540−1674.2cal=500.24cal.

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One gram of water of volume 1 c.c becomes 1671 c.c of steam when boiled at a pressure of one atmosphere. The latent heat of vaporisation of water is 540 cal/gr. Find the increase in internal energy (1 at. pressure = 105 N/m2)

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One gram of water of volume 1 c.c becomes 1671 c.c of steam when boiled at a pressure of one atmosphere. The latent heat of vaporisation of water is 540 cal/gr. Find the increase in internal energy (1 at. pressure = 10⁵ N/m²)