Question

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?

Answer 1

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"`