Question

5kg having vol of 5lit when converted into steam at ntp occupies a vol 8.405 m3 .if latent heat of vaporization of water is 540 k cal/lg .find its internal latent heat ?​

Answer 1

Answer:

Total heat energy required = change in internal energy + work done

Here, heat energy required is the latent heat of vaporization.

Change in internal energy=538−168(0.239)=538−40.152=497.848cal

( 1J=0.239cal )

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Answer 2

Answer:

497.848cal

Explanation:

• Total heat energy required= change in internal energy+ work done
• Therefore,

Change in internal energy

= 538 - 168 × 0.239

= 538- 40.152

= 497.848cal