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  • 1 cc of water at its boiling point <img alt="\mathrm{\left(100^{\circ} \mathrm{C}\right)}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Cleft%28100%5E%7B%5Ccirc%7D%20%5Cmathrm

1 cc of water at its boiling point \mathrm{\left(100^{\circ} \mathrm{C}\right)} absorbs 540 cal of heat to become steam with a volume of 1671 cc. If the atmospheric pressure is \mathrm{1.013 \times 10^5 \mathrm{Nm}^{-2}}, the energy spent (in cal) to overcome molecular attractive forces is nearly (mechanical equivalent of heat = 4.2 J/cal)

Option: 1

540


Option: 2

500


Option: 3

40


Option: 4

zero


Answers (1)

best_answer

Work done against the atmospheric pressure

                         \begin{aligned} W & =\mathrm{p \Delta V} \\ & =1.013 \times 10^5 \times(1671-1) \times 10^{-6} \mathrm{~J} \\ & =1.013 \times 167 \mathrm{~J} \\ & =\frac{1.013 \times 167}{4.2} \mathrm{cal} \cong 40 \mathrm{cal} \end{aligned}

Water absorbs 540 cal of heat out of which 40 cal used in doing work against the atmospheric pressure. Therefore, work done to overcome the molecular attraction

                               =540-40=500

Posted by

vishal kumar

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