Question

Please Help me work out his problem please and thank you.

In an electric freezer, 401.0 g of water at 18.5°C is cooled, frozen, and the ice is chilled to −5.6°C.
(a) What is the change of heat in the water?

(b) If the latent heat of vaporization of the Freon refrigerant is 40 cal/g, how many grams of Freon must be evaporated to absorb this heat

Answer 1

q1 = heat released in cooling the water from 18.5 C to zero C.
q1 = mass H2O x specific heat H2O x (Tf-Ti) = 401.0 g x cal/g*C x (0 - 18.5) = ? cakirues,
q2 = heat released on freezing the liquid H2O @ zero to solid @ zero
q2 = mass H2O x heat fusion = 401.0 g x delta H = ? You look up delta H fusion for water. Make sure the units agree with units of q1.
q3 = heat released in cooling ice from zero C to -5.6
q3 = mass ice x specific heat ice x (Tf - Ti) = 401.0 x specific heat ice x (-5.6 - 0) = ? Look up specific heat ice. Use the same units as q1 and q2. I remember specific heat ice as approx 2.09 J/g*C but that isn't calories.
Obtain total heat released by adding q1 + q2 + q3 = ? calories. Let's call this total Y calories. Then
#2. calories/grams x grams = calories
40 calories/g x What number of grams = Y calories from above.
Post your work if you get stuck.