Question

7.46 - A 0.0500-mol sample of a nutrient substance is burned in a bomb calorimeter containing 2.00 x 10g H20. If the formula weight of this nutrient substance is 114 g/mol, what is the fuel value (in nutritional Cal) if the temperature of the water increased 5.70C?​

Answer 1

Answer:

$200\frac{kcal}{g}$

Explanation:

Hello!

In this case, for this calorimetry problem, since the combustion 0.0500 mol of the nutrient increase the temperature of water by 5.70 °C, we can notice that the heat lost by the nutrient is gained by water in order to write:

$Q_{nutrient}=-Q_{water}$

Which can be also written as:

$Q_{nutrient}=-m_{water}C_{water}\Delta T_{water}\\\\Q_{nutrient}=-200 g*4.184\frac{J}{g\°C}*5.70\°C\\\\Q_{nutrient}=4769.8 J$

Thus, in terms of the grams of the nutrient:

$m_{nutrient}=0.0500mol*\frac{114g}{1mol}=5.70g$

The fuel value in nutritional Cal (kcal/g) turns out:

$Fuel \ Value=\frac{-4.7698kJ}{5.70g}*\frac{1kcal}{4.184kJ}=200\frac{kcal}{g}$

Best regards!