Consider a U-tube whose arms are open to the atmosphere. Now water is poured into the U-tube from one arm, and light oil (???? =
790 kg/m3) from the other. One arm contains 70-cm-high water, while the other arm contains both fluids with an oil-to-water height ratio of 6. Determine the height of each fluid in that arm.
1 answer:
Answer:
Oil high= 73.14cm
Water high= 12.19cm
Explanation:
See the attached images.
The first important concept is that in an open tube at both ends the atmospheric pressure balances itself.
The pressure on both sides of the tube is the same in the equilibrium. The pressure of each fluid depends on its height and density.
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Answer:
The amount of heat to absorb is 6,261 J
Explanation:
Calorimetry is in charge of measuring the amount of heat generated or lost in certain physical or chemical processes.
The total energy required is the sum of the energy to heat the ice from -20 ° C to ice of 0 ° C, melting the ice of 0 ° C in 0 ° C water and finally heating the water to 60 ° C.
So:
- Heat required to raise the temperature of ice from -20 °C to 0 °C
Being the sensible heat of a body the amount of heat received or transferred by a body when it undergoes a temperature variation (Δt) without there being a change of physical state (solid, liquid or gaseous), the expression is used:
Q = c * m * ΔT
Where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation (ΔT=Tfinal - Tinitial).
In this case, m= 10 g, specific heat of the ice= 2.1 and ΔT=0 C - (-20 C)= 20 C
Replacing: Q= 10 g*2.1 *20 C and solving: Q=420 J
- Heat required to convert 0 °C ice to 0 °C water
The heat Q necessary to melt a substance depends on its mass m and on the called latent heat of fusion of each substance:
Q= m* ΔHfusion
In this case, being 1 mol of water= 18 grams: Q= 10 g*= 3.333 kJ= 3,333 J (being kJ=1,000 J)
- Heat required to raise the temperature of water from 0 °C to 60 °C
In this case the expression used in the first step is used, but being: m= 10 g, specific heat of the water= 4.18 and ΔT=60 C - (0 C)= 60 C
Replacing: Q= 10 g*4.18 *60 C and solving: Q=2,508 J
Finally, Qtotal= 420 J + 3,333 J + 2,508 J
Qtotal= 6,261 J
<u><em> The amount of heat to absorb is 6,261 J</em></u>
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