The formula for the quantity of heat #q# transferred is
#color(blue)(bar(ul(|color(white)(a/a)q = mC_text(s)ΔTcolor(white)(a/a)|)))" "#
where
#m color(white)(ll) =# the mass of the object
#C_text(s) color(white)(l) =# its specific heat capacity
#ΔT =# its change in temperature
In this problem,
#C_text(s) color(white)(ll)= "4.184 J·°C"^"-1""g"^"-1"#
#ΔT = T_text(f) - T_text(i) = "99.0 °C - 14.5 °C = 84.5 °C"#
Your first task is to find the mass of the water.
#m = 1 color(red)(cancel(color(black)("cup"))) × (8 color(red)(cancel(color(black)("oz"))))/(1 color(red)(cancel(color(black)("cup")))) × (29.573 color(red)(cancel(color(black)("mL"))))/(1 color(red)(cancel(color(black)("oz")))) × "0.999 174 g"/(1 color(red)(cancel(color(black)("mL")))) = "236 g"#
Now you can determine the quantity of heat needed.
#q = 236 "g" × 4.184 "J"·color(red)(cancel(color(black)("°C"^"-1""g"^"-1"))) × 84.5 color(red)(cancel(color(black)("°C"))) = "83 600"color(white)(l)"J" = "83.6 kJ"#