A sample of $$0.1$$ g of water at $$l00$$°C and normal pressure.$$(1.013$$ x $$105$$ $$Nm-2$$ $$)requires$$ $$54$$ 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

Infinity Learn · Accepted Answer

$$Q=54$$ cal $$=54$$×$$4.18$$ $$joule=225.72$$ joule $$W=PVsteam-VwaterFor$$ water $$0.1$$ gram $$=$$ $$0.1$$ cc $$=0.013$$×$$105167.1$$×$$10-6-0.1$$×$$10-6joule$$ $$=1.013$$×$$167$$×$$10-1=16.917$$ joule By FLOT ⇒  ∆$$U=Q-W=225.72-16.917=208.8$$ joule.

A sample of 0.1 g of water at l00°C and normal pressure.(1.013 x 10⁵ 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:

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A sample of 0.1 g of water at l00°C and normal pressure.(1.013 x 105 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:

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Ready to Test Your Skills?

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A sample of 0.1 g of water at l00°C and normal pressure.(1.013 x 10⁵ 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: