A boy after swimming comes out from a pool covered with a film of water weighing 80 g . How much heat must be supplied to evaporate this water? (Δ Hv° = 40.79 kJ mol-1)

Question

A boy after swimming comes out from a pool covered with a film of water weighing 80 g . How much heat must be supplied to evaporate this water? (Δ Hv° = 40.79 kJ mol-1)  (A)  1.61 × 102 kJ  (B)  1.71 × 102 kJ  (C)  1.81 × 102 kJ  (D)  1.91 × 102 kJ

Tardigrade Admin · Accepted Answer

Hv = 40.79 kJ mol-1 Number of moles (n) = (80/18) Heat, q = nHv = (80/18) × 40.79 = 181.28 kJ = 1.81 × 102 kJ

Q. A boy after swimming comes out from a pool covered with a film of water weighing $80 g$ . How much heat must be supplied to evaporate this water? $(Δ H_{v}^{\circ} = 40.79 k J m o l^{- 1})$

$1.61 \times 1 0^{2} k J$

$1.71 \times 1 0^{2} k J$

$1.81 \times 1 0^{2} k J$

$1.91 \times 1 0^{2} k J$

$H_{v} = 40.79 k J m o l^{- 1}$
Number of moles $(n) = \frac{80}{18}$
Heat, $q = n H v$
$= \frac{80}{18} \times 40.79$
$= 181.28 k J$
$= 1.81 \times 1 0^{2} k J$

Q. A boy after swimming comes out from a pool covered with a film of water weighing 80g . How much heat must be supplied to evaporate this water? (ΔHv∘​=40.79kJmol−1)

1.61×102kJ

1.71×102kJ

1.81×102kJ

1.91×102kJ