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?

Nov 18, 2016

You must burn 0.026 g of the fat.

Explanation:

There are two heat transfers involved.

$\text{heat of combustion of triolein + heat gained by water = 0}$

${q}_{1} + {q}_{2} = 0$

nΔ_ cH + mcΔT = 0

In this problem,

Δ_ cH = "-3.510 × 10"^4color(white)(l) "kJ·mol"^"-1"
${M}_{r} = 885.43$

$m = \text{50 g}$
$c = \text{4.184 J°C"^"-1""g"^"-1}$
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"

${q}_{1} + {q}_{2} = \text{-3.510 × 10"^4ncolor(white)(l) "kJ" + "1.046 kJ} = 0$

n = ("-1.046" color(red)(cancel(color(black)("kJ"))))/("-3.510 × 10"^4 color(red)(cancel(color(black)("kJ")))) = 2.98 × 10^"-5"

$\text{Mass of triolein" = 2.98 × 10^"-5" color(red)(cancel(color(black)("mol triolein"))) × "885.43 g triolein"/(1 color(red)(cancel(color(black)("mol triolein")))) = "0.026 g triolein}$