Relative density of ice in cylinder

In summary, when a piece of ice is dropped into the cylinder, it sank completely in oil and the oil level rose to 90 cm3 mark. When the ice melted, the oil level came down to 87 cm3 mark. The relative density of ice is 0.98.
  • #1
leena19
186
0

Homework Statement



A measuring cylinder contains 60cm3 of oil at 0 degrees celsius .When a piece of ice is dropped into the cylinder,it sank completely in oil and the oil level rose to 90 cm3 mark.When the ice melted the oil level came down to 87cm3 mark.The relative density of ice is
(1)0.80 (2) 0.85 (3) 0.90 (4)0.95 (5)0.98

Homework Equations





The Attempt at a Solution


I guess the volume of ice is 90-60=30cm3
and the relative density of ice can be given as = real weight of ice(mg) / apparent loss in weight in oil(mg-m'g)?

mg would be 30*d*g,where d is the density of oil ,but how do I find its apparent weight in oil.I think it has something to do with the 87cm3 but I can't figure out what.

Hope someone can help.
Thanks in advance.
 
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  • #2
To calculate the relative density you need to have a reference density. For this problem I guess you're expected to calculate the relative density of ice with respect to water.

Then [itex]RD=\rho_{ice} / \rho_{water}[/itex]. Now try to find expressions for both densities.
 
  • #3
Thanx.
so [itex]\rho_{ice}[/itex] = m/30
& [itex]\rho_{water}[/itex]= m/27

therefore RD = (m/30)*(27/m) = 0.9,which would be answer no.3?
but I have a slight problem,in some questions I've used the equations
RD of the material of a body= weight of body/weight of an equal volume of water
and to find the RD of a liquid=weight of a given volume of a liquid/weight of an equal vol.of water
or upthrust in liquid/upthrust in water
I can't undersatand why we can't solve the above problem using these equations
 
  • #4
leena19 said:
but I have a slight problem,in some questions I've used the equations
RD of the material of a body= weight of body/weight of an equal volume of water

Let's use this definition then. The weight of the body of ice is [itex]W_i=m_ig=\rho_i V_i g[/itex].
The weight of an equal volume of water is [itex]W_w=m_w g=\rho_w V_i g[/itex]. Putting this in your definition of RD yields:

[tex]
RD=\frac{\rho_i V_i g}{\rho_w V_i g}=\frac{\rho_i}{\rho_w}
[/tex]

So as you can see these two equations, while perhaps looking different, really are just the same.
 
  • #5
Oh yes!They are the same.
Thanks so much,Cyosis.
 

1. What is relative density?

Relative density, also known as specific gravity, refers to the ratio of the density of a substance compared to the density of water at a specific temperature and pressure.

2. What is the relative density of ice?

The relative density of ice is approximately 0.92, meaning that it is about 92% as dense as water at the same temperature and pressure.

3. How is relative density of ice in a cylinder measured?

The relative density of ice in a cylinder can be measured by filling the cylinder with water and recording the volume. Then, the ice is added to the cylinder and the resulting volume is measured. The relative density can then be calculated by dividing the volume of the ice by the volume of water and multiplying by 100%.

4. Why is the relative density of ice important in scientific research?

The relative density of ice is important in scientific research because it affects the physical properties of water, such as its ability to float and its melting point. This can have significant implications in fields such as climate change, oceanography, and materials science.

5. How does the relative density of ice change with temperature?

The relative density of ice decreases as the temperature increases. This is due to the fact that ice expands as it freezes and becomes less dense than water. At 0°C, the relative density of ice is 1.00, meaning that it is equal in density to water at this temperature. However, as the temperature decreases below 0°C, the relative density of ice decreases as well.

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