Buoyancy
95
(M 0 ). You can see that, if the object and the weights had the same density, they
would have the same volume; each would be buoyed by the same amount, and the
correction factor would be 1.0000000 (that is, there would be no correction). If
there is a big difference in densities of weights and object (as there is in the case
of water as the object), then the correction is significant, as illustrated in the following problem.
PROBLEM:
A sample of water at 20°C is weighed in air with brass weights and found to weigh
99.8365 g. Calculate the true mass of the water. The density of brass is 8.0 g/ml and
the density of air is 1.2 x 10"' g/ml.
SOLUTION:
The buoyancy correction factor = 1 + d A \ — -- —I
\ d 0 rf u /
= 1.001050
The true weight = M 0 = A/ w (1.001050)
= (99.8365 g)( 1.001050) = 99.9413 g
' "
There is no point in using the very accurate density for water from Table 7-1
because the other densities are given only to two significant figures; furthermore,
the slight correction has almost no effect on the value of the factor or the true
weight. If we used d a = 0.9982 g/ml, the buoyancy factor would be 1.001052 and
the true weight would be 99.9415 g. The error is only two parts in a million.
A huge fraction of all weighing is done by difference — that is, the weight of
the object is found as the difference between the weight of the empty container
and the weight of the container with the object. There is no point in making a
buoyancy correction to both weighings, because the error in container weight is
the same both times and cancels out when one weight is subtracted from the
other. A buoyancy correction need be applied only to the "difference" — that is,
only to the object itself.
If the object being weighed is small, and of some significant density, the need
for buoyancy correction vanishes. The more nearly d a is likec/ u , the less important is buoyancy. For example, if you wanted to know the true weight of 0.5000
g AgCl weighed in air with brass weights (d^ = 8.0 g/ml), you would look up the
density of AgCl in a handbook (it is 5.56 g/ml) and calculate the true value as
true wtof AgCl = (0.5000 g) [l + (1.2 x 10~
3
) (j^g-- jpjj
= 0.500033 g
95
(M 0 ). You can see that, if the object and the weights had the same density, they
would have the same volume; each would be buoyed by the same amount, and the
correction factor would be 1.0000000 (that is, there would be no correction). If
there is a big difference in densities of weights and object (as there is in the case
of water as the object), then the correction is significant, as illustrated in the following problem.
PROBLEM:
A sample of water at 20°C is weighed in air with brass weights and found to weigh
99.8365 g. Calculate the true mass of the water. The density of brass is 8.0 g/ml and
the density of air is 1.2 x 10"' g/ml.
SOLUTION:
The buoyancy correction factor = 1 + d A \ — -- —I
\ d 0 rf u /
= 1.001050
The true weight = M 0 = A/ w (1.001050)
= (99.8365 g)( 1.001050) = 99.9413 g
' "
There is no point in using the very accurate density for water from Table 7-1
because the other densities are given only to two significant figures; furthermore,
the slight correction has almost no effect on the value of the factor or the true
weight. If we used d a = 0.9982 g/ml, the buoyancy factor would be 1.001052 and
the true weight would be 99.9415 g. The error is only two parts in a million.
A huge fraction of all weighing is done by difference — that is, the weight of
the object is found as the difference between the weight of the empty container
and the weight of the container with the object. There is no point in making a
buoyancy correction to both weighings, because the error in container weight is
the same both times and cancels out when one weight is subtracted from the
other. A buoyancy correction need be applied only to the "difference" — that is,
only to the object itself.
If the object being weighed is small, and of some significant density, the need
for buoyancy correction vanishes. The more nearly d a is likec/ u , the less important is buoyancy. For example, if you wanted to know the true weight of 0.5000
g AgCl weighed in air with brass weights (d^ = 8.0 g/ml), you would look up the
density of AgCl in a handbook (it is 5.56 g/ml) and calculate the true value as
true wtof AgCl = (0.5000 g) [l + (1.2 x 10~
3
) (j^g-- jpjj
= 0.500033 g