The electric stove is connected to the 220 V network. Water weighing 5 kg and a temperature of 50 °C was poured into a copper ladle weighing 800 g and placed on a tile. After 4 minutes, the water in the bucket boiled. Determine the resistance of the electric stove spiral. The loss of energy from heating the surrounding air can be neglected. The specific heat capacity of copper is 400 J/kg·°C.

This is a task ^^^

Next is the solution:

To solve the problem, we use the formula for the amount of heat required to heat water and the copper bucket:

\ [Q = mc\ Delta T\]

where:

-\ (m\) — weight (in kg),

-\ (c\) — specific heat capacity (in J/ (kg·°С)),

-\ (\ Delta T\) — temperature change (in °C).

Water mass\ (m_ {water} = 5\) kg, specific heat capacity of water\ (c_ {water} = 4200\) J/ (kg·°С), initial water temperature\ (T_ {water, start} = 50\) °С, final temperature\ (T_ {water, final} = 100\) °С.

Change in water temperature:

\ [\ Delta T_ {water} = T_ {water, of course} - T_ {water, start} = 100 - 50 = 50\] °C.

The amount of heat\ (Q_ {water}\):

\ [Q_ {water} = m_ {water}\ cdot c_ {water}\ cdot\ Delta T_ {water} = 5\ cdot 4200\ cdot 50 = 1050000\] J.

Bucket weight\ (m_ {bucket} = 0.8\) kg, specific heat capacity of copper\ (c_ {copper} = 400\) J/ (kg·°C), initial bucket temperature\ (T_ {bucket, start} = 50\) °C, final temperature\ (T_ {bucket, end} = 100\) °C.

Bucket temperature change:

\ [\ Delta T_ {bucket} = T_ {bucket, end} - T_ {bucket, start} = 100 - 50 = 50\] °C.

Amount of heat\ (Q_ {bucket}\):

\ [Q_ {bucket} = m_ {bucket}\ cdot c_ {copper}\ cdot\ Delta T_ {bucket} = 0.8\ cdot 400\ cdot 50 = 16000\] J.

The total amount of heat\ (Q_ {total}\) for heating water and the copper bucket:

\ [Q_ {total} = Q_ {water} + Q_ {bucket} = 1050000 + 16000 = 1066000\] J.

Since the total amount of heat was released in 4 minutes (240 seconds), the power\ (P\) of the electric stove:

\ [P =\ frac {Q_ {total}} {t} =\ frac {1066000} {240}\ approx 4441.67\] Watts.

Now we use Ohm's law to calculate the resistance\ (R\) of an electric stove spiral:

\ [P =\ frac {U^2} {R}\]

Where\ (U = 220\) V is the voltage.

Let's transform the formula for determining the resistance\ (R\):

\ [R =\ frac {U^2} {P} =\ frac {220^2} {4441.67}\ approach\ frac {48400} {4441.67}\ approach 10.9\] Ohm.

Thus, the resistance of the electric stove spiral is approximately\ (10.9\) Ohm.

In principle, you can understand something or tell Merlin that I'm seeing some kind of nonsense, then he'll give it without formulas that may not be displayed correctly:

Sorry for this misunderstanding. Let me present the solution to the problem in plain text without formulas that may not be displayed correctly.

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The electric stove is connected to the 220 V network. Water weighing 5 kg and a temperature of 50 °C was poured into a copper ladle weighing 800 g and placed on a tile. After 4 minutes, the water in the bucket boiled. Find the resistance of the electric stove spiral. The loss of energy from heating the surrounding air can be neglected. The specific heat capacity of copper is 400 J/ (kg·°C).

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- Water weight: 5 kg.

- Specific heat capacity of water: 4200 J/ (kg·°C).

- Initial water temperature: 50°C

- Final water temperature: 100 °C

The change in water temperature is 50 °C.

The amount of heat required to heat water:

Q_water = 5 kg 50°C = 1,050,000 J.

- Bucket weight: 0.8 kg.

- Specific heat capacity of copper: 400 J/ (kg·°C).

- Initial bucket temperature: 50°C

- Final bucket temperature: 100 °C.

The bucket temperature change is also 50°C.

The amount of heat to heat the bucket:

Q_Bucket = 0.8 kg 50°C = 16,000 J.

The total amount of heat transferred to the water and the bucket:

Q_total = Q_water+Q_bucket = 1,050,000 J+16,000 J = 1,066,000 J.

Heating time: 4 minutes, which is 240 seconds.

Electric stove power:

P = Q_total/t = 1,066,000 J/240 s = 4,441.67 W.

We use the formula:

P = U²/R,

where U = 220 V.

Let us express the resistance R:

R = U²/P = (220 V) ²/ 4,441.67 W = 48,400/ 4,441.67 ≈ 10.9 ohms.

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I mean, what the hell happened, maybe some settings flew off I don't know