55. In a fermentation process using yeast, if 500 grams of dough containing yeast rises to double its initial volume due to carbon dioxide production, what is the approximate volume increase if the initial density of the dough is 0.8 g/cm³?
Key Concept: Fermentation, Food microbiology
b) 625 cm³
[Solution Description]
First, calculate the initial volume of the dough:
$\text{Volume} = \frac{\text{Mass}}{\text{Density}} = \frac{500\,\text{g}}{0.8\,\text{g/cm}^3} = 625\,\text{cm}^3$
Since the dough doubles its volume after rising, the final volume is:
$625\,\text{cm}^3 \times 2 = 1250\,\text{cm}^3$
The volume increase is:
$1250\,\text{cm}^3 - 625\,\text{cm}^3 = 625\,\text{cm}^3$
The volume increases by 625 cm³.
Your Answer is correct.
b) 625 cm³
[Solution Description]
First, calculate the initial volume of the dough:
$\text{Volume} = \frac{\text{Mass}}{\text{Density}} = \frac{500\,\text{g}}{0.8\,\text{g/cm}^3} = 625\,\text{cm}^3$
Since the dough doubles its volume after rising, the final volume is:
$625\,\text{cm}^3 \times 2 = 1250\,\text{cm}^3$
The volume increase is:
$1250\,\text{cm}^3 - 625\,\text{cm}^3 = 625\,\text{cm}^3$
The volume increases by 625 cm³.