Question #27cd4

My guess would be that the question included the thermochemical equation that describes the formation of carbon dioxide

#"C"_ ((s)) + "O"_ (2(g)) -> "CO"_ (2(g))" " " "DeltaH_f^@ = -"393.51 kJ mol"^(-1)#

Your goal here is to figure out how much carbon, i.e. how many grams, must undergo combustion in order for the reaction to give off the amount of heat needed to heat your sample fo water.

The standard enthalpy of formation, #DeltaH_f^@#, essentially tells you how much heat is given off or absorbed when one mole of a substance is formed from its constituent elements in their most stable form.

In this case,

#DeltaH_f^@ = color(darkorange)(-"393.51 kJ")color(white)(.)color(blue)("mol"^(-1))#

tells you that when #color(blue)("1 mole"# of carbon dioxide is produced by the above reaction, #color(darkorange)("393.51 kJ")# of heat are being given off, hence the minus sign.

Now, notice that the raction consumes #1# mole of carbon to produce #1# mole of carbon dioxide, so you can say that #"393.51 kJ"# of heat ar ebeing given off when #1# mole of carbon racts to produce carbon dioxide.

You calculated the heat required to raise the temperature of #"1000 g"# of water by

#DeltaT = 40^@"C" - 10^@"C" = 30^@"C"#

by using the equation

#color(blue)(ul(color(black)(q = m * c * DeltaT)))#

and got

#q = 1000 color(red)(cancel(color(black)("g"))) * "4.18 J" color(red)(cancel(color(black)("g"^(-1)))) color(red)(cancel(color(black)(""^@"C"^(-1)))) * 30color(red)(cancel(color(black)(""^@"C")))#

#q = "125,400 J"#

Now, you know the reaction gives off #"393.51 kJ"# of heat for every mole of carbon that undergoes combustion, so use the amount of heat needed to calculate exactly how many moles would be needed here

#"125,400" color(red)(cancel(color(black)("J"))) * (1 color(red)(cancel(color(black)("kJ"))))/(10^3color(red)(cancel(color(black)("J")))) * "1 mole C"/(393.51color(red)(cancel(color(black)("kJ")))) = "318.67 moles C"#

So, in order to give off #"125,400 J"# of heat, you must burn #318.67# moles of carbon. To convert this to grams, use the molar mass of carbon

#318.67 color(red)(cancel(color(black)("moles C"))) * "12.011 g"/(1color(red)(cancel(color(black)("mole C")))) = color(darkgreen)(ul(color(black)("3800 g")))#

I'll leave the answer rounded to two sig figs, but keep in mind that your values justify only one significant figure for the answer.