Question

Допоможіть будь ласка! 1 При замикані джерела струму опором R1 = 5 Om в колі йде струм I1 = 5 A, при замикані опором R2 = 2 Om проходить струм I2 = 8 A. Знайти внутрішній опір і Е. Р. С. джерела 2 Визначити потужність нагрівника електричної каструлі, якщо в ній за час 20хв можна нагріти 20л води до температури 100 градусів Цельсія. Початкова температура води t1 = 20 градусів Цельсія К. К. Д електроприладу 70. Питома ємність води 4,2 * 10 в 3 степені Дж/кг. К

Answer 1

20хв=1200с
20л*1=20кг
Δt=100-20=80
ККД=70%=0,7
Q=c*m*Δt=4200*20*80=6720000Дж
Q=A
P=A/t*0,7=6720000/1200*0,7=8000Вт=8кВт

Answer 2

Problem 1: Finding the internal resistance and EMF of a power source

To find the internal resistance and electromotive force (EMF) of a power source, we can use the information provided about the currents and resistances in the circuit.

Given: - Resistance of R1 = 5 Ω - Current through R1 (I1) = 5 A - Resistance of R2 = 2 Ω - Current through R2 (I2) = 8 A

We can use Ohm's Law to relate the currents and resistances in the circuit:

1. For R1: - Ohm's Law: V = I * R - Voltage across R1 (V1) = I1 * R1 = 5 A * 5 Ω = 25 V

2. For R2: - Voltage across R2 (V2) = I2 * R2 = 8 A * 2 Ω = 16 V

3. Total voltage in the circuit: - The total voltage in the circuit is equal to the sum of the voltages across R1 and R2: - V_total = V1 + V2 = 25 V + 16 V = 41 V

4. Internal resistance (r) and EMF (E) of the power source: - The internal resistance (r) can be calculated using the formula: r = (V_total - V1) / I1 = (41 V - 25 V) / 5 A = 16 V / 5 A = 3.2 Ω - The EMF (E) of the power source can be calculated using the formula: E = V_total + (I2 * R2) = 41 V + (8 A * 2 Ω) = 41 V + 16 V = 57 V

Therefore, the internal resistance of the power source is 3.2 Ω and the electromotive force (EMF) of the power source is 57 V.

Problem 2: Finding the power of an electric kettle

To find the power of an electric kettle, we need to calculate the amount of energy required to heat the given amount of water from the initial temperature to the final temperature within the given time.

Given: - Initial temperature of water (t1) = 20 °C - Final temperature of water (t2) = 100 °C - Volume of water (V) = 20 L - Time taken (t) = 20 min = 20 * 60 s = 1200 s - Specific heat capacity of water (c) = 4.2 * 10^3 J/(kg·K) - Mass of water (m) = Volume * Density = V * 1000 g/L = 20 kg

1. Calculate the change in temperature: - Δt = t2 - t1 = 100 °C - 20 °C = 80 °C

2. Calculate the amount of energy required to heat the water: - Energy (Q) = m * c * Δt - Q = 20 kg * (4.2 * 10^3 J/(kg·K)) * 80 °C = 67,200,000 J

3. Calculate the power of the electric kettle: - Power (P) = Energy (Q) / Time (t) - P = 67,200,000 J / 1200 s = 56,000 W

Therefore, the power of the electric kettle is 56,000 W.

Note: The specific heat capacity of water (c) is approximately 4.2 * 10^3 J/(kg·K), and the density of water is approximately 1000 g/L.