1. The problem statement, all variables and given/known data Original Question - A fishing boat displaces 3.31×10e3 kg of liquid when floating in fresh water (ρ=1000 kg/m3). If salt water has a density of 1.05×10e3 kg/m^3, how many kilograms of water would the boat displace if it were floating in salt water? ==> answer is 3.31e3 kg My Question - By what amount would the volume of water displaced by the boat change? 3. The attempt at a solution Not sure what I'm doing wrong... d= m/v v= m/d vsalt= 3.31e4 kg/1050 kg/m^3 ~ 3.152 L voriginal = 3.31e4 kg/1000 kg/m^3 = 3.31 L Change in volume = 3.31L - 3.152 L ~0.158 L
Look at those units which you intelligently included. The volumes aren't coming out in L. They are coming out in m^3. What's the relation between the two?
The procedure is correct, but you've made a confusion with units. The boat displaces 3310 kg of water (either fresh or salt). Density of fresh water is 1000 kg/m3, so the volume displaced is 3310/1000 = 3.31 m3 (or 3310 litres) In case of salt water it's 3310/1050 = 3.152 m3 (3152 litres) The difference is 0.158 m3 or 158 litres.
I actually tried 158 L just as a guess and it was still wrong... Is that what you meant? 1 m^3 = 1000 L .158 m^3 = 158 L
If you have a coherent set of units in the input data, you better mantain it until the final result. Once you have the result, you can convert it to whatever unit sistem you prefer. For example, you have mass of water displaced in kg, and the density in kg/m3. So the result will be in (kg)/(kg/m3) = m3. Once you have the result in m3, you can apply the conversions you wrote, which are correct. If you're telling me that 158 litres is not a correct result, then I don't know where is the problem. Are the two boats on the same planet?