Question

Электрический кипятильник со спиралью сопротивлением 150 Ом поместили в сосуд и включили в сеть с напряжением 220 в. Определите на сколько градусов нагрелась вода за 5 мин?

Answer 1

Q=(U^2/R)*t=(220^2/150)*5*60=96800 Дж=96.8 кДж

Q'=c*m*dT=4.2 *m*dT=Q-->m*dT=Q/c=96.8 /4.2=23 кг*К

Для того чтобы узнать dT,надо знать массу воды.За 5 мин 23 кг нагреются на 1 градус ,а 1 кг воды на 23 градуса  и т.д.

Answer 2

Calculation of the temperature increase

To determine the temperature increase of the water, we need to consider the power dissipated by the electric kettle and the specific heat capacity of water.

The power dissipated by the electric kettle can be calculated using Ohm's Law, which states that the power (P) is equal to the square of the current (I) multiplied by the resistance (R):

P = I^2 * R Given that the resistance of the kettle is 150 Ω, we can calculate the current flowing through it. However, we don't have the current value, so we need to find it using the voltage (V) and resistance (R) values:

I = V / R Substituting the values, we get:

I = 220 V / 150 Ω

Now that we have the current, we can calculate the power dissipated by the kettle.

To calculate the temperature increase of the water, we need to know the specific heat capacity of water. The specific heat capacity of water is approximately 4.18 J/g°C.

We also need to know the mass of the water. Unfortunately, the mass of the water is not provided in the question. Therefore, we cannot calculate the exact temperature increase without this information.

However, we can provide a general formula to calculate the temperature increase based on the power dissipated by the kettle, the specific heat capacity of water, and the mass of the water:

ΔT = (P * t) / (m * c)

Where: - ΔT is the temperature increase - P is the power dissipated by the kettle - t is the time in seconds - m is the mass of the water - c is the specific heat capacity of water

Please provide the mass of the water so that we can calculate the temperature increase accurately.