Answer:
45594J
Step-by-step explanation:
Needed information in this question are;
Specific heat of ice = 2.06 J/g°C
Specific heat of water = 4.18 J/g°C
Specific heat of steam = 2.03 J/g°C
Heat of fusion of water ΔHf = 334 J/g
Melting point of water = 0 °C
Heat of vaporization of water ΔHv = 2257 J/g
Boiling point of water = 100 °C
- STEP 1:
Q1 = mcΔT
where
m = 15 grams
c (specific heat of ice) = 2.06 J/g°C
Tinitial = -5 °C
Tfinal = 0 °C
ΔT = (Tfinal – Tinitial)
ΔT = (0 °C - (-5 °C))
ΔT = 5 °C
Q1 = mcΔT
Q1 = (15 g) · (2.06 J/g°C) · (5 °C)
Q1 = 154.5 J
- STEP 2:
Q2 = m · ΔHf
where
m = 15 grams
ΔHf (heat of fusion) = 334 J/g
Q2 = m · ΔHf
Q2 = 15 · 334 J/g
Q2 = 5010 J
- STEP 3:
Q3 = mcΔT
where
m = 200 grams
c (specific heat of water) = 4.18 J/g°C
Tinitial = 0 °C
Tfinal = 100 °C
ΔT = (Tfinal – Tinitial)
ΔT = (100 °C – 0 °C)
ΔT = 100 °C
Q3 = mcΔT
Q3 = (15 g) · (4.18 J/g°C) · (100 °C)
Q3 = 6270 J
- STEP 4:
Q4 = m · ΔHv
where
m = 15 grams
ΔHv (heat of vaporization) = 2257 J/g
Q4 = m · ΔHf
Q4 = 15 · 2257 J/g
Q4 = 33855 J
- STEP 5:
Q5 = mcΔT
where
m = 15 grams
c (specific heat of steam) = 2.03 J/g°C
Tinitial = 100 °C
Tfinal = 110 °C
ΔT = (Tfinal – Tinitial)
ΔT = (110 °C – 100 °C)
ΔT = 10 °C
Q5 = mcΔT
Q5 = (15 g) · (2.03 J/g°C) · (10 °C)
Q5 = 304.5 J
Total heat = 304.5J + 33855 J + 6270 J + 5010 J + 154.5 J
= 45594J