Calculate how much work is done in supplying the heat which is
required to convert 5 gm of ice at – 2°C into steam at 100°C. (sp. heat
of ice = 0.5 cal gm-1°C- 1; L. H. of ice = 80 cal/gm and L, H. of
steam = 536 cal/gm]
Answers
Answer:
Solution:
The total energy required is the sum of the energy to heat the -10 °C ice to 0 °C ice, melting the 0 °C ice into 0 °C water, heating the water to 100 °C, converting 100 °C water to 100 °C steam and heating the steam to 150 °C. To get the final value, first calculate the individual energy values and then add them up.
Step 1: Heat required to raise the temperature of ice from -10 °C to 0 °C Use the formula
q = mcΔT
where
q = heat energy
m = mass
c = specific heat
ΔT = change in temperature
q = (25 g)x(2.09 J/g·°C)[(0 °C - -10 °C)]
q = (25 g)x(2.09 J/g·°C)x(10 °C)
q = 522.5 J
Heat required to raise the temperature of ice from -10 °C to 0 °C = 522.5 J
Step 2: Heat required to convert 0 °C ice to 0 °C water
Use the formula for heat:
q = m·ΔHf
where
q = heat energy
m = mass
ΔHf = heat of fusion
q = (25 g)x(334 J/g)
q = 8350 J
Heat required to convert 0 °C ice to 0 °C water = 8350 J
Step 3: Heat required to raise the temperature of 0 °C water to 100 °C water
q = mcΔT
q = (25 g)x(4.18 J/g·°C)[(100 °C - 0 °C)]
q = (25 g)x(4.18 J/g·°C)x(100 °C)
q = 10450 J
Heat required to raise the temperature of 0 °C water to 100 °C water = 10450 J
Step 4: Heat required to convert 100 °C water to 100 °C steam
q = m·ΔHv
where
q = heat energy
m = mass
ΔHv = heat of vaporization
q = (25 g)x(2257 J/g)
q = 56425 J
Heat required to convert 100 °C water to 100 °C steam = 56425
Step 5: Heat required to convert 100 °C steam to 150 °C steam
q = mcΔT
q = (25 g)x(2.09 J/g·°C)[(150 °C - 100 °C)]
q = (25 g)x(2.09 J/g·°C)x(50 °C)
q = 2612.5 J
Heat required to convert 100 °C steam to 150 °C steam = 2612.5
Step 6: Find total heat energy
HeatTotal = HeatStep 1 + HeatStep 2 + HeatStep 3 + HeatStep 4 + HeatStep 5
HeatTotal = 522.5 J + 8350 J + 10450 J + 56425 J + 2612.5 J
HeatTotal = 78360 J
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