A sample of 0.1 g of water at 100°C and normal pressure ($1.013 imes 10^5$ $Nm^{-2}$) requires 54 cal of heat energy to convert to steam at 100°C. If the volume of the steam produced is 167.1 cc, the change in internal energy of the sample, is

Question

A sample of 0.1 g of water at 100°C and normal pressure ($1.013 	imes 10^5$ $Nm^{-2}$) requires 54 cal of heat energy to convert to steam at 100°C. If the volume of the steam produced is 167.1 cc, the change in internal energy of the sample, is

Dr. Abubacker · Accepted Answer

[Solution] Q = 54 × 4.18 ≈ 225.7 J; W = PΔV = $1.013 	imes 10^5 	imes 167 	imes 10^{-6}$ ≈ 16.9 J; ΔU ≈ 208.7 J. Answer (2).

Q. A sample of 0.1 g of water at 100°C and normal pressure ($1.013 \times 10^5$ $Nm^{-2}$) requires 54 cal of heat energy to convert to steam at 100°C. If the volume of the steam produced is 167.1 cc, the change in internal energy of the sample, is

Explanation

Reviewed by Dr. Abubacker

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