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4 years ago

A soft drink with the properties of 10 oC water is sucked through a 4.1-mm-diameter, 0.25-m-long straw at a rate of 4.1 cm3/s.

Physics

1 answer:

nikdorinn [45]4 years ago

5 0

Answer:

1. it is laminar flow

2. it is not fully developed

3.Re=63.156

4. entrance length is the entrance diameter which is 4.1-mm-

Explanation:

A soft drink with the properties of 10 oC water is sucked through a 4.1-mm-diameter, 0.25-m-long straw at a rate of 4.1 cm3/s.

(a) Is the flow at the outlet of the straw laminar?

(b) Is it fully developed? To explain calculate

(c) the Reynolds number and

(d) the entrance length

note te following

For&pipe&flow:&

Re&<&2100 laminar&flow&

Re&>&4000 turbulent&flow&

kinematic viscosity of water at 10 oCis 1.267E-6m^2/S

Re=vl/k.viscosity

v=q/A

q=4.1 cm3/s=4.1*10^-6m3/s

A=3.142*0.0041/4

A=0.00322m2

v=0.000318m/s

l= 0.25-m

RE=0.000318m/s*0.25/( 1.267E-6m^2/S)

Re=63.156

since re<2100 , it is laminar flow

2. it is not fully developed

3.Re=63.156

4. entrance length is the entrance diameter wic is 4.1-mm-

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An aluminum calorimeter with a mass of 100 g contains 250 g of water. The calorimeter and water are in thermal equilibrium at 10

Alexeev081 [22]

Answer:

a) $c=1822.3214\ J.kg^{-1}.K^{-1}$

b) This value of specific heat is close to the specific heat of ice at -40° C and the specific heat of peat (a variety of coal).

c) The material is peat, possibly.

d) The material cannot be ice because ice doesn't exists at a temperature of 100°C.

Explanation:

Given:

mass of aluminium, $m_a=0.1\ kg$

mass of water, $m_w=0.25\ kg$

initial temperature of the system, $T_i=10^{\circ}C$

mass of copper block, $m_c=0.1\ kg$

temperature of copper block, $T_c=50^{\circ}C$

mass of the other block, $m=0.07\ kg$

temperature of the other block, $T=100^{\circ}C$

final equilibrium temperature, $T_f=20^{\circ}C$

We have,

specific heat of aluminium, $c_a=910\ J.kg^{-1}.K^{-1}$

specific heat of copper, $c_c=390\ J.kg^{-1}.K^{-1}$

specific heat of water, $c_w=4186\ J.kg^{-1}.K^{-1}$

Using the heat energy conservation equation.

The heat absorbed by the system of the calorie-meter to reach the final temperature.

$Q_{in}=m_a.c_a.(T_f-T_i)+m_w.c_w.(T_f-T_i)$

$Q_{in}=0.1\times 910\times (20-10)+0.25\times 4186\times (20-10)$

$Q_{in}=11375\ J$

The heat released by the blocks when dipped into water:

$Q_{out}=m_c.c_c.(T_c-T_f)+m.c.(T-T_f)$

where

$c=$ specific heat of the unknown material

For the conservation of energy : $Q_{in}=Q_{out}$

so,

$11375=0.1\times 390\times (50-20)+0.07\times c\times (100-20)$

$c=1822.3214\ J.kg^{-1}.K^{-1}$

This value of specific heat is close to the specific heat of ice at -40° C and the specific heat of peat (a variety of coal).

The material is peat, possibly.

The material cannot be ice because ice doesn't exists at a temperature of 100°C.

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