Question

a piece of gold aluminum alloy weighs 490 N in air, when suspended from a spring balance and submerged in water, the balance reads 360 N. what is the weight of the gold in the alloy if the sg of gold is 19.3 and the sg of aluminum is 2.7. answer is 570 N . (pls. i need help :( for the solution ) thanks

Answer 1

Did not solve completely.

volume of sample, V, is same as for displaced water,
which provided 490-360=130N buoyancy
which would be provided by 130/9.8 = 13.3 kg =.0133 m3 of water = V.

In air, weight was 490N so mass was 490/9.8 = 50 kg.
Its density is thus 50 kg / .0133 m3 = 3760 kg/m3, a specific gravity of 3.76

Let X = mass fraction gold and 1-X be mass fraction of aluminum, then they ust combine to give the sp. gr. of the sample:

19.3 X + 2.7 (1-X) = 3.76

16.6 X =1.06

X = 0.064 gold and 1-X = 0.936 aluminum mass fractions

weight of gold is (0.064) (50 kg) (9.8 N/kg) = 31.4 N

well, that does not match "the answer," which must be wrong because you cannot have more gold weight than you have sample weight.