The formula for the heat #q# gained or lost by a substance is
#color(blue)(bar(ul(|color(white)(a/a)q = mcΔTcolor(white)(a/a)|)))" "#
where
#m# is the mass of the substance.
#c# is its specific heat capacity.
#ΔT = T_"f" - T_"i"# is the change in temperature.
In this problem,
#q = "5.78 kJ" = "5780 J"#
#m = "328 g"#
#C = "4.18 J°C"^"-1""g"^"-1"#
#ΔT = T_"f"color(white)(l) "- 35.3 °C"#
#q = 328 color(red)(cancel(color(black)("g"))) × 4.18 color(red)(cancel(color(black)("J")))"°C"^"-1"color(red)(cancel(color(black)("g"^"-1"))) × ΔT = 5780 color(red)(cancel(color(black)("J")))#
#1371ΔT "°C"^"-1" = 5780#
#ΔT = 5780/(1371 "°C"^"-1") = "4.22 °C"#
#ΔT = T_"f"color(white)(l) "- 35.3 °C" = "4.22 °C"#
#T_"f" = "35.3 °C + 4.22 °C" = "39.5 °C"#
