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?

Question

Serenity Allen · Accepted Answer

You must burn 0.026 g of the fat.

There are two different heat transfers at work. #"heat of combustion of triolein + heat gained by water = 0"# #q_1 + q_2 = 0# #nΔ_ cH + mcΔT = 0# In this issue, #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"#

Xavier Avery · Answer

undefined Use the heat equation: $ q = mc\Delta T $, where $ q $ is heat, $ m $ is mass, $ c $ is specific heat, and $ \Delta T $ is temperature change. Convert kJ to J and find the mass of glyceryl trioleate.