1 cm 3 of water at its boiling point absorbs 540 cal of heat to become steam with a volume of 1671 cm 3 the atmospheric pressure =1.013 × 105 Nm -2 and the mechanical equivalent of heat =4.19 cal, the energy spent in this process in overcoming intermolecular forces is:

Question

1 cm 3 of water at its boiling point absorbs 540 cal of heat to become steam with a volume of 1671 cm 3 the atmospheric pressure =1.013 × 105 Nm -2 and the mechanical equivalent of heat =4.19 cal, the energy spent in this process in overcoming intermolecular forces is: (A) 500 cal (B) 499 cal (C) 502 cal (D) 490 cal

Tardigrade Admin · Accepted Answer

Work done by steam = P Δ V =1.013 × 105[1671-1] × 10-6 =169.17 Joule =41.06 cal ∴ [1 cal =4.2 Joule ] energy spent to overcoming intermolecular forces is =540-41.06=498.94 cal ≃ 499 cal

$500 c a l$

$499 c a l$

$502 c a l$

$490 c a l$

Work done by steam $= P Δ V$
$= 1.013 \times 1 0^{5} [1671 - 1] \times 1 0^{- 6}$
$= 169.17$ Joule
$= 41.06$ cal $∴ [1$ cal $= 4.2$ Joule $]$
energy spent to overcoming intermolecular forces is
$= 540 - 41.06 = 498.94 c a l ≃ 499 c a l$

Q. 1cm3 of water at its boiling point absorbs 540 cal of heat to become steam with a volume of 1671cm3 the atmospheric pressure =1.013×105Nm−2 and the mechanical equivalent of heat =4.19 cal, the energy spent in this process in overcoming intermolecular forces is:

500cal

499cal

502cal

490cal

Work done by steam =PΔV =1.013×105[1671−1]×10−6 =169.17 Joule =41.06 cal ∴[1 cal =4.2 Joule ] energy spent to overcoming intermolecular forces is =540−41.06=498.94cal≃499cal

$500 c a l$

$499 c a l$

$502 c a l$

$490 c a l$

Work done by steam $= P Δ V$
$= 1.013 \times 1 0^{5} [1671 - 1] \times 1 0^{- 6}$
$= 169.17$ Joule
$= 41.06$ cal $∴ [1$ cal $= 4.2$ Joule $]$
energy spent to overcoming intermolecular forces is
$= 540 - 41.06 = 498.94 c a l ≃ 499 c a l$