Two grams o liquid water are at 0 ° C, and another two grams of liquid water are

Question

Two grams o liquid water are at 0° C, and another two grams of liquid water are at 100° C. Heat is removed from the water at 0° C, completely freezing it at 0° C. This heat is then used to vaporize some of the water at 100° C. The latent heats of fusion and vaorization of water at 1 atm are 3.35 x 105 J/kg and 2.26 x 106 J/kg, respectively.

(4) What is the amount of heat removed from the water at 0° C?

A) 470 J; B) 510 J; C) 570 J; D) 620 J; E) 670 J.

(5) What is the mass of the water that remains a liquid at 100° C?

A) 1.40 g; B) 1.50 g; C) 1.60 g; D) 1.70 g; E) 1.80 g.

Explanation / Answer

4) the amount of heat removed from the water at 0° C is

dQ = m * L(fusion)

dQ = 0.002 * 3.35* 10^5

dQ = 670 J

Answer E) 670 J

5) this dQ = m * L(vaporization)

m = 670/2.26* 10^6

m = 0.3 g

the mass of the water that remains a liquid at 100° C

is 2-0.3 = 1.79g

Answer D) 1.70 g

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