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Chapter 1: Problem 39

The information on a can of pop indicates that the can contains \(355 \mathrm{mL}\). The mass of a full can of pop is \(0.369 \mathrm{kg}\), while an empty can weighs 0.153 N. Determine the specific weight, density, and specific gravity of the pop and compare your results with the corresponding values for water at \(20^{\circ} \mathrm{C}\). Express your results in SI units.

Short Answer

Expert verified
The specific weight, density, and specific gravity of the pop are calculated based on the weight, mass, and volume of the pop alone, and compared with values for water at \(20^{\circ}C\). The calculations would reveal whether the pop is more or less dense and heavy than water, and by how much.

Step by step solution

01

Finding the Volume of Pop

The volume of the pop is indicated on the can as 355 mL. To convert this to cubic meters (m^3), which is the SI unit for volume, use the conversion factor \(1 mL = 1 \times 10^{-6} m^{3}\). Therefore, the volume of the pop, V, is \(355 \times 10^{-6} m^{3}\).
02

Calculating the Weight and Mass of Pop

The mass of the full can is given as 0.369 kg and the weight of the empty can as 0.153 N. To find the mass of just the pop, subtract the mass of the empty can (which can be found by dividing its weight by the acceleration due to gravity, 9.81 ms^-2), from the mass of the full can. The resultant M is the mass of the pop. To find the weight of the pop, multiply this mass by 9.81 ms^-2.
03

Calculating Specific Weight, Density, and Specific Gravity

The specific weight (also called the weight density) is the weight of pop per unit volume, calculated by dividing the weight of pop by the volume of pop. The density is the mass of the pop per unit volume, calculated by dividing the mass of pop by the volume of pop. Specific gravity is the ratio of the density of the pop to the density of water, which is \(1 x 10^{3} kg/m^{3}\) at \(20^{\circ}C\).
04

Comparing with Values for Water

Specific weight, density, and specific gravity of water at \(20^{\circ}C\) are 9.81 kN/m^3, \(1 x 10^{3} kg/m^{3}\), and 1, respectively. The calculated values for pop should be compared to these.

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Most popular questions from this chapter

1.17 A formula to estimate the volume rate of flow, \(Q\), flowing over a dam of length, \(B\), is given by the equation \\[ Q=3.09 B H^{3 / 2} \\] where \(H\) is the depth of the water above the top of the dam (called the head). This formula gives \(Q\) in \(\mathrm{ft}^{3} / \mathrm{s}\) when \(B\) and \(H\) are in feet. Is the constant, \(3.09,\) dimensionless? Would this equation be valid if units other than feet and seconds were used?

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