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postnew [5]
1 year ago
14

Water in a metal pot is heated over a campfire. Primarily, by wat means of energy transfer is thermal energy moved from the fire

to the bottom of the pot?
Physics
1 answer:
Art [367]1 year ago
7 0

when water in a metal pot is heated over a campfire. Primarily, by means of convection energy transfer is thermal energy moved from the fire to the bottom of the pot

<h3 /><h3>What is thermal energy?</h3>

It can be defined as the form of the energy in which heat is transferred from one body to another body due to their molecular movements, thermal energy is also known as heat energy.

Thermal energy can flow by different means such as conduction, convection, and radiation.

As given in the problem when water in a metal pot is heated over a campfire. The energy we receive around the campfire is mostly due to the effect of the radiation of the heat but the Primary means of energy transfer is thermal energy moved from the fire to the bottom of the pot because of the convection.

Thus, when water is heated in a metal pot over a campfire. Thermal energy is primarily transferred from the fire to the bottom of the pot through convection.

Learn more about thermal energy from here

brainly.com/question/3022807

#SPJ1

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Air enters a nozzle steadily at 2.21 kg/m3 and 20 m/s and leaves at 0.762 kg/m3 and 150 m/s. If the inlet area of the nozzle is
saveliy_v [14]

Answer:

a) The mass flow rate through the nozzle is 0.27 kg/s.

b) The exit area of the nozzle is 23.6 cm².

Explanation:

a) The mass flow rate through the nozzle can be calculated with the following equation:

\dot{m_{i}} = \rho_{i} v_{i}A_{i}

Where:

v_{i}: is the initial velocity = 20 m/s

A_{i}: is the inlet area of the nozzle = 60 cm²  

\rho_{i}: is the density of entrance = 2.21 kg/m³

\dot{m} = \rho_{i} v_{i}A_{i} = 2.21 \frac{kg}{m^{3}}*20 \frac{m}{s}*60 cm^{2}*\frac{1 m^{2}}{(100 cm)^{2}} = 0.27 kg/s  

Hence, the mass flow rate through the nozzle is 0.27 kg/s.

b) The exit area of the nozzle can be found with the Continuity equation:

\rho_{i} v_{i}A_{i} = \rho_{f} v_{f}A_{f}

0.27 kg/s = 0.762 kg/m^{3}*150 m/s*A_{f}

A_{f} = \frac{0.27 kg/s}{0.762 kg/m^{3}*150 m/s} = 0.00236 m^{2}*\frac{(100 cm)^{2}}{1 m^{2}} = 23.6 cm^{2}

Therefore, the exit area of the nozzle is 23.6 cm².

I hope it helps you!                                                                  

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