Apuddle 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 = 540 cal/g)
a. +1.8 c°
b. -1.8 c°
c. +0.18 c°
d. −0.18 c°

The correct answer is b

Explanation:

In this exercise we must assume that no heat from the environment enters, the initial heat of the puddle is distributed in the heat of the evaporated water and the heat of liquid water remaining

Let's look for the heat to evaporate the water

      Q₁ = m L

      Q₁ = 0.50 540

      Q₁ = 270 cal

The remaining water is

     m = 150 -0.50 = 149.5 gr

The heat for this water is

     Q2 = m ce DT

The amount of heat should be conserved, the heat assigned is equal to minus the heat absorbed

      Q₁ = - Q₂

      Q₁ = -m ce ΔT

      ΔT = -Q₁ / m ce

      Δt = -270 / (149.5 1)

      ΔT = -1.8 ° C

The water has cooled 1.8 ° c

The correct answer is b

a high pressure system is a dense air mass that is usually cooler and drier than the surrounding air. a low pressure system is a less dense air mass that is usually wetter and warmer than the surrounding air.

in general, areas that experience high atmospheric pressure also experience fair weather. low pressure systems can cause the formation of clouds and storms. air usually flows from areas of high pressure to areas of low pressure..

are you like autistic or something?

xd