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4 years ago

13

How much heat energy, in kilojoules, is required to convert 53.0 g of ice at −18.0 ∘C to water at 25.0 ∘C ? Express your answer

to three significant figures and include the appropriate units.

1 answer:

ANTONII [103]4 years ago

8 0

Answer:

25.2 kJ of heat energy is required to convert 53.0 g of ice at −18.0 ∘C to water at 25.0 ∘C

Explanation:

Data: mass of ice = m = 53.0 g

Temperature of ice = T1 = -18.0 ∘C

Temperature of water = T2 = 25.0 ∘C

Change in Temperature = ΔT : T2-T1

Specific heat of ice = c(i) = 2.09 J/g . ∘C

Specific heat of water = c(w) = 4.18 J/g . ∘C

Enthalpy of fusion of water (when convert from ice to water) = ΔH(f) : 334 J/g

Enthalpy of vapourization of water (when convert from water to gas) = ΔH(v) :2250 J/g

Total heat required = q = ?

Solution: Heat required to melt the ice

T1 = -18 ∘C

T2 = 0 ∘C

Q1 = m x c(i) x ΔT

Q1 = 53.0 x 2.09 x (0-(-18))

Q1 = 53.0 x 2.09 x 18

Q1 = 1993.86 J is required by ice to reach to its

melting point.

<u>Heat required to convert the ice into water</u>

Q2 = ΔH(f) x m

Q2 = 334 x 53.0 = 17702 J is required by ice to

convert into water.

<u> Heat required by water to reach at 25 ∘C</u>

T1 = 0 ∘C

T2 = 25 ∘C

Q3 = m x c(w) x ΔT

Q3 = 53.0 x 4.18 x (25-0)

Q3 = 5538.5 J is required by water to reach at

25∘C .

<u>Total heat required by 53.0 g of ice at −18.0 ∘C </u>

<u>to water at </u>

<u>25.0 ∘C</u>

q = Q1 + Q2 + Q3

q = 1993.86 J + 17702 J + 5538.5 J

q = 25234.36 J is the total heat required 53.0 g

of ice at −18.0 ∘C to water at 25.0 ∘C.

<u>Conversion from Joule (J) to Kilojoule (kJ)</u>

1000 J = 1 kJ

q = 25234.36 J/1000

q = 25.23436 kJ

<u>Conversion of heat in kJ to three significant </u>

<u>figures</u>

q = 25.2 kJ

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