This is a disguised calorimetry problem, with two heat transfers involved.
#color(white)(mmmm)"NH"_4"NO"_3"(s)" color(white)(mmml)→ color(white)(m)"NH"_4"NO"_3"(aq)"#
#"heat of solution of NH"_4"NO"_3 + "heat lost by water" = 0#
#color(white)(mmmmml)q_1 color(white)(mmmmmmm)+color(white)(mmmm) q_2color(white)(mmmll) = 0#
#color(white)(mmmll)nΔ_text(soln)Hcolor(white)(mmmmmll) + color(white)(mmll)mCΔT = 0#
In this problem,
#n = 500 color(red)(cancel(color(black)("g NH"_4"NO"_3))) × (1 "mol NH"_4"NO"_3)/(80.04 color(red)(cancel(color(black)("g NH"_4"NO"_3)))) = "6.247 mol NH"_4"NO"_3"#
#Δ_text(soln)H = "25.41 kJ· mol"^"−1"#
∴ #q_1 = 6.247 color(red)(cancel(color(black)("mol"))) × "25 410 J"·color(red)(cancel(color(black)("mol"^"-1"))) = "158 700 J"#
I am guessing that you are using US gallons. Then
#m = 1.50 color(red)(cancel(color(black)("gal"))) × (3.785 color(red)(cancel(color(black)("L"))))/(1 color(red)(cancel(color(black)("gal")))) × ("1000 g")/(1 color(red)(cancel(color(black)("L")))) = "5678 g"#
#C = "4.184 J·g"^"-1""°C"^"-1"#
∴ # q_2 = 5678 color(red)(cancel(color(black)("g"))) × "4.184 J"·color(red)(cancel(color(black)("g"^"-1")))"°C"^"-1" × ΔT = "23 750 J·°C"^"-1"·ΔT#
#q_1 + q_2 = "158 700" color(red)(cancel(color(black)("J"))) + "23 750" color(red)(cancel(color(black)("J")))"°C"^"-1"·ΔT = 0#
#ΔT = "-158 700"/"23 750 °C"^"-1" = "-6.68 °C"#
