Question

Электрический чайник питается от напряжения 220 В при токе 3,2 А.Определить КПД(%) чайника,если 1 л воды с начальной температурой 20 С закипает через 12 минут. Св= 4200 Дж/кг*С

Answer 1

Мощность тока:

P = U*I = 220*3,2 = 704 Вт

Количество теплоты на нагревание воды:

Q = c * m * (t2 - t1)

Имеем :
с = 4200 Дж/кг*K
m = 1 кг

t2 = 100 градусов
t1 = 20 градусов

Q = 4200*1*(100-20) = 336 000 Дж = 336 кДж

Работа тока:

А = N*t, где t=12 мин = 12*60 = 720 с
А = 704*720 = 506 880 Дж = 506,8 кДж
 Коэффициент полезного действия чайника:

КПД = Q*100%/A = 336*100/506,8 = 66,3 %

Ответ: КПД = 66,3 %

Answer 2

Calculation of Efficiency of the Electric Kettle

To determine the efficiency of the electric kettle, we need to calculate the amount of energy used to heat the water and compare it to the electrical energy consumed by the kettle.

Given information: - Voltage (V) = 220 V - Current (I) = 3.2 A - Time (t) = 12 minutes = 720 seconds - Specific heat capacity of water (Sv) = 4200 J/kg°C - Initial temperature (t1) = 20°C

Step 1: Calculate the electrical energy consumed by the kettle

The electrical energy consumed by the kettle can be calculated using the formula:

Electrical energy (E) = Voltage (V) x Current (I) x Time (t)

Substituting the given values: E = 220 V x 3.2 A x 720 s

Calculating the value of E: E = 506880 J

Step 2: Calculate the energy required to heat the water

The energy required to heat the water can be calculated using the formula:

Energy (Q) = Mass (m) x Specific heat capacity (Sv) x Change in temperature (ΔT)

Given: - Mass of water (m) = 1 kg - Initial temperature (t1) = 20°C - Final temperature (t2) = boiling point of water = 100°C

Substituting the values: Q = 1 kg x 4200 J/kg°C x (100°C - 20°C)

Calculating the value of Q: Q = 1 kg x 4200 J/kg°C x 80°C Q = 336,000 J

Step 3: Calculate the efficiency of the electric kettle

The efficiency (η) of the electric kettle can be calculated using the formula:

Efficiency (η) = (Energy output / Energy input) x 100%

Substituting the values: η = (Q / E) x 100%

Calculating the value of η: η = (336,000 J / 506,880 J) x 100% η ≈ 66.3%

Therefore, the efficiency of the electric kettle is approximately 66.3%.

Please note that the above calculation assumes ideal conditions and does not take into account any energy losses due to factors such as heat dissipation or electrical resistance.