a tungsten and coil has a resistance of 12 ohm at 15 degree celsius is the temperature coefficient of resistance of tungsten is 0.004 calculate the equal resistance at 80 degree Celsius​

Answers

Answer 1
Answer:

Answer:

The resistance of the tungsten coil at 80 degrees Celsius is 15.12 ohm

Explanation:

The given parameters are;

The resistance of the tungsten coil at 15 degrees Celsius = 12 ohm

The temperature coefficient of resistance of tungsten = 0.004/°C

The resistance of the tungsten coil at 80 degrees Celsius is found using the following relation;

R₂ = R₁·[1 + α·(t₂ - t₁)]

Where;

R₁ = The resistance at the initial  temperature = 12 ohm

R₂ = The resistance of tungsten at the final temperature

t₁ = The initial temperature = 15 degrees Celsius

t₂ = The final temperature = 80 degrees Celsius

α = temperature coefficient of resistance of tungsten = 0.004/°C

Therefore, we have;

R₂ = 12×[1 + 0.004×(80 - 15)] = 15.12 ohm

The resistance of the tungsten coil at 80 degrees Celsius = 15.12 ohm.


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Answers

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hope it helps

Answer:

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Answers

number 2 pretty sure..

Consider a room that is initially at the outdoor temperature of 20°C. The room contains a 40-W light bulb, a 110-W TV set, a 300-W refrigerator, and a 1110-W iron. Assuming no heat transfer through the walls, determine the rate of increase of the energy content of the room when all of these electric devices are on.

Answers

Heat transfer is the flow of heat due to temperature difference.

The rate of increase in the energy content of the room is 1560 W.

What is heat transfer?

Heat transfer is the flow of heat due to temperature difference. The heat flow from higher temperature to the lower temperature.

Given information-

Initial temperature of the room is 20 degree Celsius.

The room contains a 40-W light bulb, a 110-W TV set, a 300-W refrigerator, and a 1110-W iron.

The rate of increase in the energy of the room is the difference of energy increasing in the room and the energy leaving to the room.

Thus the rate of increase

(dE_(room))/(dt)=E_(in)-E_(out)

Energy inside the room is the sum of all the energies increasing  as,

\rm E_(in)=E_(light)+E_(refrigerator)+E_(TV)+E_(iron)

As there is no heat transfer through the walls. Thus the energy leaving to the room is,

E_(out)=0

Thus put the values in the first equation as,

(dE_(room))/(dt)=E_(in)-E_(out)\n(dE_(room))/(dt)=40++110+300+1110\n(dE_(room))/(dt)=1560

Thus the rate of increase in the energy content of the room is 1560 W.

Learn more about the heat transfer here;

brainly.com/question/12072129

Final answer:

The rate of increase of the energy content of the room when all electric devices are on is 1560 W.

Explanation:

To determine the rate of increase of the energy content of the room, we need to calculate the total power consumption of the electric devices. Adding up the power ratings, we have:



Total power = 40 W + 110 W + 300 W + 1110 W = 1560 W



Therefore, the rate of increase of the energy content of the room when all of these electric devices are on is 1560 W.

Learn more about Power consumption here:

brainly.com/question/36790528

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Answers

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