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
