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A block of rock with edge length 8.5cm, 9.0 cm, 4.3-8cm was a mass of 953kg, find the Specific gravity of the rock

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3.22\text{3.22}

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Question 5:
Given: Mass of rock M=953 kgM = 953 \, \text{kg}, Length L=85.5 cmL = 85.5 \, \text{cm}, Width W=79.0 cmW = 79.0 \, \text{cm}, Height H=43.8 cmH = 43.8 \, \text{cm}.
Density of water ρw=1000 kg/m3\rho_w = 1000 \, \text{kg/m}^3.

  1. Convert the dimensions of the rock from centimeters to meters.

    L=85.5 cm=0.855 mL = 85.5 \, \text{cm} = 0.855 \, \text{m} W=79.0 cm=0.790 mW = 79.0 \, \text{cm} = 0.790 \, \text{m} H=43.8 cm=0.438 mH = 43.8 \, \text{cm} = 0.438 \, \text{m}

  2. Calculate the volume of the rock (VrockV_{rock}).

    Vrock=L×W×HV_{rock} = L \times W \times H Vrock=0.855 m×0.790 m×0.438 mV_{rock} = 0.855 \, \text{m} \times 0.790 \, \text{m} \times 0.438 \, \text{m} Vrock≈0.2960 m3V_{rock} \approx 0.2960 \, \text{m}^3

  3. Calculate the density of the rock (ρrock\rho_{rock}).

    ρrock=MVrock\rho_{rock} = \frac{M}{V_{rock}} ρrock=953 kg0.2960 m3\rho_{rock} = \frac{953 \, \text{kg}}{0.2960 \, \text{m}^3} ρrock≈3219.59 kg/m3\rho_{rock} \approx 3219.59 \, \text{kg/m}^3

  4. Calculate the specific gravity (GsG_s) of the rock.

    Gs=ρrockρwG_s = \frac{\rho_{rock}}{\rho_w} Gs=3219.59 kg/m31000 kg/m3G_s = \frac{3219.59 \, \text{kg/m}^3}{1000 \, \text{kg/m}^3} Gs≈3.22G_s \approx 3.22

    The specific gravity of the rock is 3.22\text{3.22}

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