The formula for the heat qq 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"
