Question

Электрокипятильник со спиралью сопротивлением 160 Ом поместили в сосуд, содержащий 0,5 л воды при 20 °С, и включили в сеть напряжением 220 В. Через 20 мин спираль выключили. Какое количество воды выкипело, если КПД спирали 80%?Учитель сказала решать через эти формулы: Q=Y^2*R*t Y=U/RQ1=c*m*(t2-t1) Q3=h*mПомогите пожалуйста. В интернете все не правильно.

Answer 1

Электрокипятильник со спиралью сопротивлением R=160 Ом поместили в сосуд, содержащий V=0,5 л воды при t1=20 °С, и включили в сеть напряжением U=220 В. Через t=20 мин=1200 c спираль выключили. Какое количество воды выкипело, если КПД спирали  КПД=80%?  M- ?

КПД= Qп/Qз
Qз=U^2*t/R=220*220*1200/160=363000 Дж
0,8*Qз=с*m*Δt + L*M

M=(0,8*Qз-с*m*Δt)/L=(0.8*363000-4200*0,5*80)/2,3*10^6=272*10^-3 кг=272 г

Ответ M=272 г

проверь вычисления

Answer 2

Calculating the Amount of Water Vaporized

To calculate the amount of water vaporized, we can use the formula for heat energy:

Q = Y^2 * R * t

Where: - Q is the heat energy absorbed by the water (in joules) - Y is the current flowing through the resistor (in amperes) - R is the resistance of the resistor (in ohms) - t is the time for which the current flows (in seconds)

Given: - Resistance (R) = 160 ohms - Voltage (U) = 220 V - Time (t) = 20 minutes = 1200 seconds - Efficiency (η) = 80%

First, we need to calculate the current (Y) using Ohm's law: U = Y * R Y = U / R

Then, we can calculate the heat energy (Q) absorbed by the water using the formula: Q = Y^2 * R * t

After finding the heat energy absorbed, we can use the specific heat capacity of water to find the amount of water vaporized.

Let's calculate step by step.

Calculating Current (Y)

Using Ohm's law: Y = U / R Y = 220 V / 160 Ω Y ≈ 1.375 A

Calculating Heat Energy (Q)

Using the formula: Q = Y^2 * R * t Q = (1.375 A)^2 * 160 Ω * 1200 s Q ≈ 2640 J

Calculating Amount of Water Vaporized

The heat energy (Q) absorbed by the water is 2640 J. Given that the efficiency (η) is 80%, we can calculate the actual heat energy used to vaporize the water: Actual Q = Q / η Actual Q = 2640 J / 0.80 Actual Q = 3300 J

Now, we can use the specific heat capacity of water (c) to find the amount of water vaporized: Q1 = c * m * ΔT Where: - c is the specific heat capacity of water (4186 J/kg°C) - m is the mass of water vaporized (in kg) - ΔT is the change in temperature (in °C)

Given: - Specific heat capacity of water (c) = 4186 J/kg°C - Change in temperature (ΔT) = the boiling point of water (100°C) - initial temperature (20°C) = 80°C

We can rearrange the formula to solve for the mass of water vaporized (m): m = Q1 / (c * ΔT) m = 3300 J / (4186 J/kg°C * 80°C) m ≈ 0.099 kg

So, approximately 0.099 kg of water vaporized during the 20 minutes the coil was active.