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allsm [11]
3 years ago
5

I have a special thermos that has a vacuum chamber between my hot soup and its outside wall. Because my soup is isolated from it

s surrounding by a vacuum,
A. My soup will stay warm forever because thermal energy cannot pass through a vacuum.
B. My soup will cool down because of conduction.
C. My soup will cool down because of radiation.
D. My soup will cool down because of convection.E. Both B and D.
Physics
1 answer:
bagirrra123 [75]3 years ago
7 0

Answer:

C. My soup will cool down because of radiation.

Explanation:

Since, there is vacuum between hot soup and its outside wall, then heat can not flow through conduction and convection.

The heat then only flows through radiation.

Therefore, the soup will not cool down because of convection or conduction by because of radiation.

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sleet_krkn [62]

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4 0
3 years ago
A small water pump is used in an irrigation system. The pump takes water in from a river at 10oC, 100 kPa at a rate of 5 kg/s. T
sergij07 [2.7K]

Answer:

0.98kW

Explanation:

The conservation of energy is given by the following equation,

\Delta U = Q-W

\dot{m}(h_1+\frac{1}{2}V_1^2+gz_1)-\dot{W} = \dot{m}(h_2+\frac{1}{2}V_2^2+gz_)

Where

\dot{m} = Mass flow

h_1 =Specific Enthalpy (IN)

h_2 = Specific Enthalpy (OUT)

g = Gravity

z_{1,2} = Heigth state (In, OUT)

V_{1,2} =Velocity (In, Out)

Our values are given by,

T_i = 10\°C

P_1 = 100kPa

\dot{m} = 5kg/s

z_2 = 20m

For this problem we know that as pressure, temperature as velocity remains constant, then

h_1 = h_2

V_1 = V_2

Then we have that our equation now is,

\dot{m}(gz_1) = \dot{m}(gz_2)+\dot{W}

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\dot{W} = -0.98kW

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