To determine the total weight of a 1 cm thick salt deposit covering the horizontal passage and the Queen's Chamber (excluding its floor) of the Great Pyramid of Khufu, we must logically break down the problem into geometric surface area calculations and volumetric mass conversions based on the density of sea salt.

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To solve this step by step, we first establish the dimensions of the spaces involved, calculate their relevant surface areas, multiply by the 1 cm thickness to find the volume, and finally apply the density of sea salt to find the total weight.

Step 1: Dimensions and Surface Area of the Queen's Chamber (Excluding Floor)

The Queen's Chamber in the Great Pyramid of Khufu is roughly rectangular with a gabled roof. Based on standard archaeological measurements (such as those by Flinders Petrie or Morton Edgar):

• Length (East-West): $\approx 5.23$ meters ($523\mathrm{\text{cm}}$)
• Width (North-South): $\approx 5.74$ meters ($574\mathrm{\text{cm}}$)
• Wall Height: $\approx 4.57$ meters ($457\mathrm{\text{cm}}$) at the peak of the gabled roof, with side walls rising to about $4.67$ meters. For simplicity, we use the standard flat-equivalent wall and gabled ceiling geometry.
• Floor Area to Exclude: $\mathrm{\text{Length}}\times \mathrm{\text{Width}}=5.23\mathrm{\text{m}}\times 5.74\mathrm{\text{m}}\approx 30.02{\mathrm{\text{m}}}^2$ ($300,200{\mathrm{\text{cm}}}^2$).
• Total Surface Area of Walls and Ceiling:
  • North and South walls: $2\times (\mathrm{\text{Width}}\times \mathrm{\text{Height}})=2\times (5.74\mathrm{\text{m}}\times 4.57\mathrm{\text{m}})\approx 52.45{\mathrm{\text{m}}}^2$
  • East and West walls: $2\times (\mathrm{\text{Length}}\times \mathrm{\text{Height}})=2\times (5.23\mathrm{\text{m}}\times 4.57\mathrm{\text{m}})\approx 47.78{\mathrm{\text{m}}}^2$
  • Gabled Ceiling: Composed of two slanted slabs meeting at the ridge. With a span of $5.74\mathrm{\text{m}}$ and a pitch, the total two-sided ceiling area is approximately $31.50{\mathrm{\text{m}}}^2$.
  • Total Queen's Chamber Target Area: $\approx 52.45+47.78+31.50=131.73{\mathrm{\text{m}}}^2$ ($1,317,300{\mathrm{\text{cm}}}^2$).

Step 2: Dimensions and Surface Area of the Horizontal Passage

The horizontal passage leading to the Queen's Chamber features a lowered section and a main horizontal corridor:

• Length: $\approx 35.2$ meters ($3,520\mathrm{\text{cm}}$)
• Width: $\approx 1.05$ meters ($105\mathrm{\text{cm}}$)
• Height: $\approx 1.20$ meters ($120\mathrm{\text{cm}}$)
• Since the deposit covers the entire horizontal passage (walls, ceiling, and floor, unlike the chamber where the floor is excluded), we calculate the total inner perimeter multiplied by the length.
• Perimeter: $2\times (\mathrm{\text{Width}}+\mathrm{\text{Height}})=2\times (1.05\mathrm{\text{m}}+1.20\mathrm{\text{m}})=4.50\mathrm{\text{m}}$.
• Total Horizontal Passage Surface Area: $\mathrm{\text{Perimeter}}\times \mathrm{\text{Length}}=4.50\mathrm{\text{m}}\times 35.2\mathrm{\text{m}}\approx 158.40{\mathrm{\text{m}}}^2$ ($1,584,000{\mathrm{\text{cm}}}^2$).

Step 3: Total Volume of the Salt Deposit

• Combined Surface Area: $131.73{\mathrm{\text{m}}}^2+158.40{\mathrm{\text{m}}}^2=290.13{\mathrm{\text{m}}}^2$
• Thickness of Deposit: $1\mathrm{\text{cm}}=0.01\mathrm{\text{m}}$
• Total Volume ($V$): $\mathrm{\text{Surface Area}}\times \mathrm{\text{Thickness}}=290.13{\mathrm{\text{m}}}^2\times 0.01\mathrm{\text{m}}=$ $2.9013{\mathrm{\text{m}}}^3$ (or $2,901,300{\mathrm{\text{cm}}}^3$).

Step 4: Weight Calculation

The bulk density of loose or compacted sea salt ranges typically between $1.0{\mathrm{\text{g/cm}}}^3$ to $1.3{\mathrm{\text{g/cm}}}^3$, with solid crystal sea salt or compressed bulk salt averaging around $2.16{\mathrm{\text{g/cm}}}^3$ ($2,160{\mathrm{\text{kg/m}}}^3$). Using the standard crystal density for 100% sea salt:

• $\mathrm{\text{Weight}}=\mathrm{\text{Volume}}\times \mathrm{\text{Density}}$
• $\mathrm{\text{Weight}}=2.9013{\mathrm{\text{m}}}^3\times 2,160{\mathrm{\text{kg/m}}}^3\approx$ $6,266.8\mathrm{\text{kg}}$ (or roughly $6.27\mathrm{\text{metric tonnes}}$).

Thus, a 1 cm thick layer of 100% sea salt across these specified surfaces would weigh approximately $6,267\mathrm{\text{kg}}$.[1]