Question

A rectangular block of solid carbon (graphite) floats at the interface of two immiscible liquids.The bottom liquid is a relatively heavy lubricating oil, and the top liquid is water. Of the total block volume, 59.2% is immersed in the oil and the balance is in the water.In a separate experiment, an empty flask is weighed, 35.3 cm3 of the lubricating oil is poured into the flask, and the flask is reweighed.If the scale reading was 124.8 g in the first weighing, what would it be in the second weighing?

Answer 1

Answer:

Second reading : 235.2 g

Explanation:

Given:

Volume of oil displaced  V_oil = 0.592*V_b

Volume of water displaced V_water = 0.408*V_b

Density of graphite p_g = 2.26 g / cm^3

Density of water p_water = 1 g / cm^3

Find: mass of oil

Step 1: Density of oil (p_oil)

p_g * V_b = p_oil * V_oil + p_water * V_water

p_g * V_b = p_oil * 0.592*V_b + p_water * 0.408*V_b

p_g = 0.592*p_oil + 0.408*p_water

p_oil =  (p_g - 0.408*p_water) / 0.592

p_oil = 3.1284 g / cm^3

Step 2: Mass of oil in flask

m_oil = p_oil * V_oil

m_oil = 3.1284 * 35.3

m_oil = 110.4325 g

Step 3: Second weigh scale reading

Second reading = m_oil + First reading

Second reading = 110.4325 + 124.8

Second reading = 235.2 g

Answer 2

Answer:

The scale reading on second weighing is 233.5g.

The solution to this problem is dependent on the density of the graphite block used. In this solution 2230kg/m³ was used.

The full details of the solution can be found in the attachment below.

Explanation: