Question

Электрический кипятильник с сопративлением 160 ом поместили в сосуд содержащий 0.5 л. воды при 20 градусах по цельсию и включили в сеть с напряжением 220 вольт.Через какое время вода закипит если кпд=80%.

Answer 1

Дано                       Си                    Решение
R= 160  Ом                               КПД= Ап\ Аз
V=0,5 л              0,0005 м^3      Ап- работа полезная нагреть воду
р=1000 кг/м^3                            Аз- работа затраченнная связана с током
t1= 20  C                                      Ап= Q= cm ( t2-t1)          m=pV= 0,5 кг
t2=100  C                                     Q= 4200 * 0,5* 80=168000 дж
U=220 B                                      Аз= U^2  T /  R
КПД= 80%      0,8                        T= Аз * R /  U^2

Найти : Т                                    Аз= Ап .\ Кпд= 168000\ 0,8=210 000
                                                        Т= 210 000 * 160 /  220^2=     694c              

Answer 2

Calculation of Time for Water to Boil

To calculate the time it takes for the water to boil, we need to consider the power of the electric kettle and the efficiency of the heating process. The power of the electric kettle can be calculated using the formula:

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

In this case, the voltage is 220 volts. However, we don't have information about the current. Therefore, we cannot calculate the power directly. However, we are given the efficiency (η) of the electric kettle, which is 80%. The efficiency is defined as the ratio of the useful output energy to the input energy.

The useful output energy in this case is the heat energy transferred to the water, and the input energy is the electrical energy consumed by the kettle. The relationship between power, time, and energy can be expressed as:

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

Since we don't have the power, we can rewrite the equation as:

Energy (E) = (Voltage (V) * Current (I)) * Time (t)

We can rearrange the equation to solve for time:

Time (t) = Energy (E) / (Voltage (V) * Current (I))

To calculate the energy transferred to the water, we can use the specific heat capacity of water, which is approximately 4.18 J/g°C. The energy can be calculated using the formula:

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

In this case, the mass of water is 0.5 kg (since 1 liter of water is approximately equal to 1 kg), the specific heat capacity is 4.18 J/g°C, and the temperature change is the difference between the boiling point of water (100°C) and the initial temperature (20°C).

Now let's calculate the time it takes for the water to boil.

Calculation Steps:

1. Calculate the energy transferred to the water: - Mass (m) = 0.5 kg - Specific Heat Capacity (c) = 4.18 J/g°C - Temperature Change (ΔT) = 100°C - 20°C = 80°C - Energy (E) = (0.5 kg) * (4.18 J/g°C) * (80°C)

2. Calculate the time: - Voltage (V) = 220 volts - Efficiency (η) = 80% = 0.8 - Current (I) = Power (P) / Voltage (V) - Power (P) = Efficiency (η) * Voltage (V) * Current (I) - Time (t) = Energy (E) / (Voltage (V) * Current (I))

Let's perform the calculations.

Calculation:

1. Calculate the energy transferred to the water: - Energy (E) = (0.5 kg) * (4.18 J/g°C) * (80°C) = 1672 J

2. Calculate the time: - Power (P) = (0.8) * (220 volts) * Current (I) - Time (t) = (1672 J) / ((220 volts) * Current (I))

Since we don't have the value of the current, we cannot calculate the exact time it takes for the water to boil. The current value is required to determine the power of the electric kettle and, consequently, the time it takes for the water to reach its boiling point.

Please provide the value of the current so that we can calculate the time accurately.