Question

1. The metabolism of one mole of glycerol trioleate, C.,H,O, a common fat, produces 3.022 x 10ºkJ of heat. How many grams of the fat must be burned to raise the temperature of 50 g of water from 25.0°C to 30.0°C?"

Answer 1

Given:

m = 50 g

C = 4.184 J/g.oC

Ti = 25 oC

Tf = 30 oC

use:

Q = m*C*(Tf-Ti)

Q = 50.0*4.184*(30.0-25.0)

Q = 1046 J

Q = 1.046 KJ

This is the amount of heat required.

Mol of C51H93O6 required = amount of heat required / heat supplied by one mol

= 1.046 KJ / (3.022*10^4 )

= 3.46*10^-5 mol

Molar mass of C51H93O6,

MM = 51*MM(C) + 93*MM(H) + 6*MM(O)

= 51*12.01 + 93*1.008 + 6*16.0

= 802.254 g/mol

use:

mass of C51H93O6,

m = number of mol * molar mass

= 3.46*10^-5 mol * 8.023*10^2 g/mol

= 2.776*10^-2 g

Answer: 2.78*10^-2 g