find total energy, which you need, to completely transform 130g of ice of temper
ID: 1783062 • Letter: F
Question
find total energy, which you need, to completely transform 130g of ice of temperature 0°C to a steam. For H2O the heat of melting is 335kJ/kg (or 80 cal/g) and the heat of evaporation is 2261 kJ/kg (or 540 kcal/kg), and specific heat capacity for H2O c=4.2 kJ/kg°K (or 1 kcal/kg°C). Water steams standard T=100°C find total energy, which you need, to completely transform 130g of ice of temperature 0°C to a steam. For H2O the heat of melting is 335kJ/kg (or 80 cal/g) and the heat of evaporation is 2261 kJ/kg (or 540 kcal/kg), and specific heat capacity for H2O c=4.2 kJ/kg°K (or 1 kcal/kg°C). Water steams standard T=100°C find total energy, which you need, to completely transform 130g of ice of temperature 0°C to a steam. For H2O the heat of melting is 335kJ/kg (or 80 cal/g) and the heat of evaporation is 2261 kJ/kg (or 540 kcal/kg), and specific heat capacity for H2O c=4.2 kJ/kg°K (or 1 kcal/kg°C). Water steams standard T=100°CExplanation / Answer
Useful information:
heat of fusion of water = 335 J/g
heat of vaporization of water = 2261 J/g
specific heat of ice = 2.09 J/g·°C
specific heat of water = 4.2 J/g·°C
specific heat of steam = 2.09 J/g·°C
Step 1) Heat required to convert 0 °C ice to 0 °C water
Use the formula
q = m·Hf
where
q = heat energy
m = mass
Hf = heat of fusion
q = (130 g)x(335 J/g)
q = 43550 J
Heat required to convert 0 °C ice to 0 °C water = 43550 J
Step 2) Heat required to raise the temperature of 0 °C water to 100 °C water
q = mcT
q = (130 g)x(4.2 J/g·°C)[(100 °C - 0 °C)]
q = (130 g)x(4.2 J/g·°C)x(100 °C)
q = 54600 J
Heat required to raise the temperature of 0 °C water to 100 °C water = 54600 J
Step 3) Heat required to convert 100 °C water to 100 °C steam
q = m·Hv
where
q = heat energy
m = mass
Hv = heat of vaporization
q = (130 g)x(2261 J/g)
q = 293930 J
Heat required to convert 100 °C water to 100 °C steam = 293930 J
Step 4) Find total heat energy
HeatTotal = HeatStep 1 + HeatStep 2 + HeatStep 3
HeatTotal = 43550 + 54600 + 293930 = 392080 J
Answer:
The heat required to convert 130 grams of 0 °C ice into 100 °C steam is 392080 J or 392.08 kJ.
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