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

b mirrors submit a iisner? e. for which optical devices does d sometimes have a positive value? o a a and b o b. a and c o c. b and d 0 d. b and cb

massxgravity x heght=weightxheight= pe

4900=490xh

h=10