In: Chemistry
how much heat (in kj) is needed to convert 866 g of ice at -10 c to steam at 126 c?
spH of ice and steam are 2.03J/g*c and 1.99J/g*c, respectively.
See, the total energy needed for this conversion 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 126 °C(this is
because of phase transitions that take place while converting from
ice to steam.)
we will do in different steps.
1) Heat required to raise the temperature of ice
from -10 °C to 0 °C
we will use the formula:
q = mcΔT
where
q = heat energy
m = mass
c = specific heat
ΔT = change in temperature
q = (866 g)x(2.03 J/g·°C)[(0 °C - (-10 °C)]
q = (866 g)x(2.03 J/g·°C)x(10 °C)
q = 17579.8 J
Heat required to raise the temperature of ice from -10 °C to 0 °C =
17579.7 J = 17.5798 kJ
2) Heat required to convert 0 °C ice to 0 °C
water
Using the formula
q = m·ΔHf
where
q = heat energy
m = mass
ΔHf = heat of fusion
q = (866 g)x(334 J/g) (heat of fusion taken from google)
q = 289244 J = 289.244 kJ
Heat required to convert 0 °C ice to 0 °C water = 289.244 kJ
3) Heat required to raise the temperature of 0 °C
water to 100 °C water
q = mcΔT
q = (866 g)x(4.18 J/g·°C)[(100 °C - 0 °C)]
q = (866 g)x(4.18 J/g·°C)x(100 °C)
q = 361988 J
Heat required to raise the temperature of 0 °C water to 100 °C
water = 361988 J= 361.988 kJ
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 = (866 g)x(2257 J/g)
q = 1954562 J
Heat required to convert 100 °C water to 100 °C steam = 1954.562
kJ
Step 5: Heat required to convert 100 °C steam to
126 °C steam
q = mcΔT
q = (866 g)x(1.99 J/g·°C)[(126 °C - 100 °C)]
q = (866 g)x(1.99 J/g·°C)x( 26°C)
q = 44806.84 J
Heat required to convert 100 °C steam to 126 °C steam = 44.80684
kJ
Step 6: Find total heat energy
HeatTotal = HeatStep 1 + HeatStep 2 + HeatStep 3 + HeatStep 4 +
HeatStep 5
HeatTotal = 2668180.54 J = 2668.18054 kJ