There are two heat transfers involved.
#"heat of combustion of triolein + heat gained by water = 0"#
#q_1 + q_2 = 0#
#nΔ_ cH + mcΔT = 0#
In this problem,
#Δ_ cH = "-3.510 × 10"^4color(white)(l) "kJ·mol"^"-1"#
#M_r = 885.43#
#m = "50 g"#
#c = "4.184 J°C"^"-1""g"^"-1"#
#ΔT = T_f - T_i = "30.0 °C - 25.0 °C" = "5.0 °C"#
#q_1 = nΔ_cH = n color(red)(cancel(color(black)("mol"))) × ("-3.510 × 10"^4color(white)(l) "kJ"·color(red)(cancel(color(black)("mol"^"-1"))))= "-3.510 × 10"^4ncolor(white)(l) "kJ"#
#q_2 = mcΔT = 50 color(red)(cancel(color(black)("g"))) × "4.184 J"·color(red)(cancel(color(black)("°C"^"-1""g"^"-1"))) × 5.0 color(red)(cancel(color(black)("°C"))) = "1046 J" = "1.046 kJ"#
#q_1 + q_2 = "-3.510 × 10"^4ncolor(white)(l) "kJ" + "1.046 kJ" = 0#
#n = ("-1.046" color(red)(cancel(color(black)("kJ"))))/("-3.510 × 10"^4 color(red)(cancel(color(black)("kJ")))) = 2.98 × 10^"-5"#
#"Mass of triolein" = 2.98 × 10^"-5" color(red)(cancel(color(black)("mol triolein"))) × "885.43 g triolein"/(1 color(red)(cancel(color(black)("mol triolein")))) = "0.026 g triolein"#