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Solnce55 [7]
3 years ago
15

cup of hot chocolate has a spoon in it. How does the heat move from the hot chocolate to the metal spoon handle?

Physics
1 answer:
serg [7]3 years ago
5 0

The heat moves from the hot chocolate to the handle of the spoon by a process called thermal conduction.

It is the transfer of heat energy from one object to another when they are in contact with eachother.

Hope this answers your question.

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samantha a high school teacher wants to spend more time with her children and is looking for a job in her field. what would be a
sp2606 [1]

She could look for an online job or a job she can work at from home.

3 0
2 years ago
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what percentage of North American adults may be functioning below their potential due to prolonged exposure to stress
Yuliya22 [10]
I believe the percentage is between 15-20%. Stress is a well known factor that affects the performance of people.
6 0
3 years ago
efrigerant-134a is expanded isentropically from 600 kPa and 70°C at the inlet of a steady-flow turbine to 100 kPa at the outlet.
PolarNik [594]

Answer:

Inlet : v_i=0.0646\frac{m}{s}

Outlet:  v_o=0.171\frac{m}{s}

Explanation:

1) Notation and important concepts

Flow of mass represent "the mass of a substance which passes per unit of time".

Flow rate represent "a measure of the volume of liquid that moves in a certain amount of time"

Specific volume is "the ratio of the substance's volume to its mass. It is the reciprocal of density."

Isentropic process is a "thermodynamic process, in which the entropy of the fluid or gas remains constant".

We know that the flow of mass is given by the following expression

\dot{m}=\frac{\dot{V}}{\upsilon}, where \dot{V} represent the flow rate and \upsilon the specific volume at the pressure and temperature given.

A_i=0.5m^2 is the inlet area

P_i=600Kpa pressure at the inlet area

T_i=70C temperature at the inlet area

A_o=1m^2 is the outlet area

P_o=100Kpa pressure at the outlet area

T_o=C temperature at the outlet area

\dot{m}=0.75\frac{kg}{s} represent the flow of mass

If we look at the first figure attached Table A-13 we see that the specific volume for the inlet condition is

\upsilon_i =0.04304\frac{kg}{m^3} and the entropy is h_i=1.0645\frac{KJ}{KgK}=h_o

With the value of entropy and the outlet pressure of 100 Kpa we can find we specific volume at the outlet condition since w ehave the entropy h_o=1.0645\frac{KJ}{KgK}

Since on the table we don't have the exact value we need to interpolate between these two values (see the second figure attached)

h_1=1.0531\frac{KJ}{KgK} , \upsilon_1=0.22473\frac{kg}{m^3}

h_2=1.0829\frac{KJ}{KgK} , \upsilon_2=0.23349\frac{kg}{m^3}

Our interest value would be given using interpolation like this:

\upsilon=0.22473+\frac{(0.23349-0.22473)}{(1.0829-1.0531)}(1.0645-1.0531)=0.228\frac{kg}{m^3}

2) Solution to the problem

Now since we have all the info required to solve the problem we can find the velocities on this way.

We know from the definition of flow of mass that \dot{m}=\frac{\dot{V}}{\upsilon}, but since \dot{V}=Av we have this:

\dot{m}=\frac{Av}{\upsilon}

If we solve from the velocity v we have this:

v=\frac{\upsilon \dot{m}}{A}   (*)

And now we just need to replace the values into equation (*)

For the inlet case:

v_i=\frac{\upsilon_i \dot{m}}{A_i}=\frac{0.043069\frac{kg}{m^3}(0.75\frac{kg}{s})}{0.5m^2}=0.0646\frac{m}{s}

For the oulet case:

v_o=\frac{\upsilon_o \dot{m}}{A_o}=\frac{0.228\frac{kg}{m^3}(0.75\frac{kg}{s})}{1m^2}=0.171\frac{m}{s}

7 0
3 years ago
Two streams merge to form a river. One stream has a width of 8.4 m, depth of 3.5 m, and current speed of 2.2 m/s. The other stre
hoa [83]

Answer:

Explanation:

We shall solve this problem on the basis of pinciple that water is incompressible so volume of flow will be equal at every point .

rate of volume flow of one stream

= cross sectional area x velocity

= 8.4 x 3.5 x 2.2 = 64.68 m³ /s

rate of volume flow of other stream

= 6.6 x 3.6 x 2.7

= 64.15 m³ /s

rate of volume flow of rive , if d be its depth

= 11.2 x d x 2.8

= 31.36 d

volume flow of river = Total of volume flow rate of two streams

31.36 d  = 64.15 + 64.68

31.36 d  = 128.83

d = 4.10 m /s .

6 0
2 years ago
A book is to be produced with pages of thickness 0.125mm.If the book is to be exactly 1cm thick what is the maximum number of pa
miskamm [114]

Answer:

1cm=10mm

10cm/0.125mm

80pages

5 0
2 years ago
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