How many grams of propane must be burned to provide the heat needed to melt a 70.0-g piece of ice...?
How many grams of propane must be burned to provide the heat needed to melt a 70.0-g piece of ice (at its melting point) and bring the resulting water to a boil (with no vapourization)? Assume a constant pressure of 1 bar and use the following data:
ΔHfo(H2O(g))=-241.83 kJ/mol, ΔHfo(O2(g))=0 kJ/mol, ΔHfo(CO2(g))=-393.5 kJ/mol, ΔHfo(C3H8(g))=-103.8 kJ/mol
Tfus(H2O)=0 oC, Tvap(H2O)=100 oC
when I tried doing it I got 0.63 but I don't think that's correct
This is what I did:
moles of H2O = 3.88
delta H rxn = -2044.02
(3.88 moles)(6.01 kJ/mol) = 23.359 kJ
(3.88 moles)(75.291 J/mol/C) = 29.263 kJ
23.359kJ + 29.2636 kJ = 29.286 kJ
(44 g of C3H8)(29.286kJ/-2044.02kJ) = 0.6304 g
But I think I did something wrong??
To calculate the grams of propane required to provide the heat needed to melt a 70.0-g piece of ice and bring the resulting water to a boil, we need to consider two processes: the melting of ice and the heating of water.
First, let's calculate the amount of heat required to melt the ice. The heat required for this process can be calculated using the formula:
Q = m * ΔHfus,
where Q is the heat required, m is the mass of the ice, and ΔHfus is the heat of fusion of water.
Given:
m = 70.0 g (mass of the ice)
ΔHfus = ΔHfo(H2O) = -241.83 kJ/mol (standard heat of fusion of water)
To convert the kJ/mol value into J/g, we need to divide it by the molar mass of water (18.02 g/mol).
Thus, ΔHfus = -241.83 kJ/mol / 18.02 g/mol = -13.41 kJ/g.
Now, substitute the values into the formula:
Q = m * ΔHfus = 70.0 g * (-13.41 kJ/g) = -938.7 kJ.
The negative sign indicates the energy absorbed during the phase change.
Next, we need to calculate the amount of heat required to heat the resulting water from its melting point (0°C) to its boiling point (100°C). The formula for this calculation is:
Q = m * C * ΔT,
where Q is the heat required, m is the mass of the water, C is the specific heat capacity of water, and ΔT is the change in temperature.
Given:
m = 70.0 g (mass of the water)
C = 4.18 J/g°C (specific heat capacity of water)
ΔT = (100 - 0) °C = 100°C (change in temperature)
Substituting the values:
Q = m * C * ΔT = 70.0 g * 4.18 J/g°C * 100°C = 293,260 J.
Now, let's convert the total energy needed from joules to kilojoules:
Q = 293,260 J = 293.26 kJ.
Finally, we add the two quantities of heat together to find the total energy required:
Total Q = Q(melting ice) + Q(heating water) = -938.7 kJ + 293.26 kJ = -645.44 kJ.
So, we need -645.44 kJ of energy to melt the ice and heat the water.
To find the grams of propane required, we can use the equation:
ΔHcomb = Q = m * ΔHcomb,
where ΔHcomb is the heat of combustion of propane.
Given:
ΔHcomb = ΔHfo(C3H8(g)) = -103.8 kJ/mol (standard heat of combustion of propane).
Again, we need to convert the kJ/mol value to J/g by dividing it by the molar mass of propane (44.1 g/mol).
Thus, ΔHcomb = -103.8 kJ/mol / 44.1 g/mol = -2.35 kJ/g.
Now, let's rearrange the equation to solve for the mass of propane:
m = Q / ΔHcomb = -645.44 kJ / (-2.35 kJ/g) = 275.17 g.
Therefore, approximately 275.17 grams of propane must be burned to provide the heat required to melt a 70.0-g piece of ice and bring the resulting water to a boil.