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Question

A bread contains 40% (by volume) edible matter and the remaining space is filled with air. If the density of edible matter is 2 g/cc, what will be the bulk density of the bread (in g/cc)?

The correct answer is
0.8

Understanding Bread Bulk Density Calculation

The bulk density of an object like bread considers its total mass spread over its total volume, including any pore spaces filled with air.

Calculating Components of Bread

  • Assume a total volume for the bread, for simplicity, let's use $V_{total} = 1 \text{ cc}$.
  • The volume of edible matter is 40% of the total volume: $V_{matter} = 0.40 \times V_{total} = 0.40 \text{ cc}$.
  • The remaining volume is filled with air: $V_{air} = (1 - 0.40) \times V_{total} = 0.60 \times V_{total} = 0.60 \text{ cc}$.
  • The density of the edible matter is given as $\rho_{matter} = 2 \text{ g/cc}$.
  • The density of air is negligible ($\rho_{air} \approx 0 \text{ g/cc}$) and does not contribute significantly to the mass.

Determining Total Mass and Bulk Density

  • Calculate the mass of the edible matter: $m_{matter} = V_{matter} \times \rho_{matter} = 0.40 \text{ cc} \times 2 \text{ g/cc} = 0.8 \text{ g}$.
  • The mass of the air is effectively zero: $m_{air} = V_{air} \times \rho_{air} \approx 0.60 \text{ cc} \times 0 \text{ g/cc} = 0 \text{ g}$.
  • The total mass of the bread is the sum of the mass of the edible matter and air: $m_{total} = m_{matter} + m_{air} = 0.8 \text{ g} + 0 \text{ g} = 0.8 \text{ g}$.
  • The bulk density ($\rho_{bulk}$) is the total mass divided by the total volume: $\rho_{bulk} = \frac{m_{total}}{V_{total}} = \frac{0.8 \text{ g}}{1 \text{ cc}} = 0.8 \text{ g/cc}$.

Therefore, the bulk density of the bread is 0.8 g/cc.

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