A puddle holds 150 g of water. If 0.50 g of water evaporates from the surface, what is the approximate temperature change of the remaining water? (Lv = 2259 × 10^3 J/kg)

1 Answer
A08
Mar 7, 2018

Water in puddle #=150/1000=0.15\ kg#

Water evaporated #0.50/1000=5.0xx10^-4\ kg#
For evaporation this water will use latent heat of evaporation of water.
Heat required for evaporation of this amount of water #Q=mL_v#

#Q=5.0xx10^-4xx2259 × 10^3=1.1295xx10^3\ J#

This heat will be taken from the remaining water in the puddle. Which is given by

#Q=msDeltaT#
where #m# is mass of water remaining in the puddle, #s=4186\ Jkg^-1K^-1# is specific heat of water and #DeltaT# is the change of temperature of water in the puddle.

Equating both and inserting various values we get

#(0.15-5.0xx10^-4)xx4186xxDeltaT=1.1295xx10^3#
#=>DeltaT=(1.1295xx10^3)/((0.15-5.0xx10^-4)xx4186)#
#=>DeltaT=1.8^@\C#

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