The metabolism of one mole of glyceryl trioleate, #C_57H_104O_6#, a common fat, produces #3.510 ×10^4# kJ of heat. How many grams of the fat must be burned to raise the temperature of 50 g of water from 25.0C to 30.0C?

1 Answer
Nov 18, 2016

Answer:

You must burn 0.026 g of the fat.

Explanation:

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"#