What is the final temperature that the water in a bomb calorimeter reaches if #DeltaE_(rxn) = -"2882 kJ/mol"# for the combustion of #"1 mol"# of butane, and the initial temperature was #20.13^@ "C"#?

#T_f = 65.94^@ "C"#

This should make sense, because the water should have been heated.

The idea is that the combustion of butane inside the bomb calorimeter releases heat out into the water to heat the water.

You were given #DeltaE_(rxn) = -"2882 kJ/mol butane"#, and so, for a small mass of butane, you can calculate the heat produced.

From the first law of thermodynamics,

#DeltaE = q + w#

where #q# is the heat flow and #w# is the compression/expansion work.

But a bomb calorimeter is rigid (has constant volume), so #w = 0# and

#DeltaE = q_V#

where #q_V# is the heat flow at constant volume for some conservative thermodynamic process.

Next, the amount of butane in #mols# is:

#10.09 cancel("g C"_4"H"_10) xx ("1 mol C"_4"H"_10)/(4 xx 12.011 cancel"g" + 10 xx 1.0079 cancel"g") = "0.1736 mols butane"#

As a result, the heat released from the reaction was:

#q_(V,rxn) = DeltaE_(rxn) = -"2882 kJ"/cancel"mol butane" xx 0.1736 cancel"mols butane" = -"500.3 kJ"#

Then, this amount of heat is presumed to heat the water, so

#-q_(V,rxn) = q_(V,soln) = m_wC_wDeltaT_w#,

where #m_w# is the mass of water in #"g"# and #C_w# is its specific heat capacity (#"4.184 J/g"cdot"K"#).

Therefore:

#q_(V,soln) = "500300 J"#

#= "2610 g water" cdot "4.184 J/g"cdot"K" cdot (T_f - 20.13^@ "C")#

Now you can solve for the final temperature.

#color(blue)(T_f) = ("500300 J")/("2610 g" cdot "4.184 J/g"cdot"K") + 20.13^@ "C"#

#= color(blue)(65.94^@ "C")#