Question

Электрический кипятильник за 9 минут 14 секунд нагревает один кг воды от 20 °C До кипения определите сопротивление нагревательного элемента кипятильником, по которому протекает ток силой 5 А, если КПД кипятильника 80 % решите пж с дано со всеми формулами и с решением

Answer 1

дано:

t = 554c;

m = 1кг;

∆t = 80°C;

i = 5A;

ŋ = 80% = 0,8.

найти: R - ?

решение:

ŋ = Nпол./Nзатр.

Nпол. = i²/R.

Nзатр. = Q/t =cm∆t/t.

ŋ = i²/R : cm∆t/t = i²t/Rcm∆t.

ŋRcm∆t = i²t;

R = i²t/ŋcm∆t;

R = 5²*554/0,8*4200*1*80 = 13850/268800 = 0,051 Ом.

ответ: R = 0,051 Oм.

Answer 2

Problem Analysis

We are given the following information: - The electric kettle heats 1 kg of water from 20 °C to boiling in 9 minutes and 14 seconds. - The efficiency of the kettle is 80%. - The current passing through the heating element is 5 A.

We need to determine the resistance of the heating element.

Solution

To solve this problem, we can use the formula for power:

Power (P) = Current (I) * Voltage (V)

Since we know the current (I) and the efficiency (η), we can calculate the power (P) using the formula:

Power (P) = I^2 * R / η

where R is the resistance of the heating element.

We also know that the power is equal to the energy transferred per unit time:

Power (P) = Energy (E) / Time (t)

The energy transferred can be calculated using the formula:

Energy (E) = Mass (m) * Specific Heat Capacity (c) * Temperature Change (ΔT)

where: - Mass (m) is 1 kg, - Specific Heat Capacity (c) of water is 4186 J/kg°C, - Temperature Change (ΔT) is the difference between the final and initial temperatures.

We can rearrange the formula for power to solve for resistance:

R = (P * η) / I^2

Let's calculate the resistance using the given values.

Calculation

Given: - Mass (m) = 1 kg - Specific Heat Capacity (c) = 4186 J/kg°C - Initial Temperature (T1) = 20 °C - Final Temperature (T2) = 100 °C (boiling point of water) - Time (t) = 9 minutes and 14 seconds = 554 seconds - Current (I) = 5 A - Efficiency (η) = 80%

First, let's calculate the energy transferred (E):

ΔT = T2 - T1 = 100 °C - 20 °C = 80 °C

E = m * c * ΔT = 1 kg * 4186 J/kg°C * 80 °C = 334,880 J

Next, let's calculate the power (P):

P = E / t = 334,880 J / 554 s = 604.33 W

Finally, let's calculate the resistance (R):

R = (P * η) / I^2 = (604.33 W * 0.8) / (5 A)^2 = 96.69 Ω

Therefore, the resistance of the heating element is approximately 96.69 Ω.

Answer

The resistance of the heating element in the kettle, through which a current of 5 A flows, is approximately 96.69 Ω.