Asked by Lucy
The heat of combustion of bituminous coal is 2.5 x 10^4 J/g. What quantity of coal is required to produce the energy required to convert 100. Pounds of ice at 0.0 degrees Celsius to steam at 100.0 degrees Celsius?
Answers
Answered by
DrBob222
First, convert 100 lbs ice to grams. 100 lbs x (454 g/lb) = ?grams. You may want to use 453.6 g/lb or 453.59 g/lb..
Second, determine q needed to heat the ice & water.
q1 = heat needed to melt ice.
q1 = mass ice x heat fusion.
q2 = heat needed to raise liquid water from zero C to 100 C.
q2 = mass water x specific heat H2O x (Tfinal-Tinitial)
q3 = heat needed to convert water at 100 to steam at 100.
q3 = mass H2O x heat vaporization.
Total q = q1 + q2 + q3.
Then,
2.54E4 J/g x #g coal = q to heat H2O.
Substitute and solve for #g coal.
Post your work if you get stuck.
Second, determine q needed to heat the ice & water.
q1 = heat needed to melt ice.
q1 = mass ice x heat fusion.
q2 = heat needed to raise liquid water from zero C to 100 C.
q2 = mass water x specific heat H2O x (Tfinal-Tinitial)
q3 = heat needed to convert water at 100 to steam at 100.
q3 = mass H2O x heat vaporization.
Total q = q1 + q2 + q3.
Then,
2.54E4 J/g x #g coal = q to heat H2O.
Substitute and solve for #g coal.
Post your work if you get stuck.
Answered by
Lucy
Does an answer around 5.36563 kg sound about right? After adding q1+q2+q3 I got 136287 kj so I converted that to J and then found the #g coal to be 5.36563 kg.
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