Question

Помогите с задачей по физике! (желательно дано и решение) Двигатель мощностью 50 Вт в течение 10 мин вращает лопасти вентилятора внутри калориметра с водой. На сколько повысится за это время температура воды, если ее масса 2 кг? Считайте, что работа вентилятора затрачивается только на нагревание воды. ДАЮ 70 БАЛЛОВ ЗА РЕШЕНИЕ!

Answer 1

Найдем работу вентилятора A= P*t
по условию   A=Q
P*t=c*m*Δt

Δt=P*t/c*m=50*600/4200*2=3,57 С

Answer 2

Problem Statement

We have a fan inside a calorimeter filled with water. The fan is powered by a 50 W motor and rotates the blades for 10 minutes. We need to calculate the increase in temperature of the water inside the calorimeter, given that the mass of the water is 2 kg. We can assume that all the work done by the fan is used to heat the water.

Solution

To solve this problem, we need to calculate the amount of heat transferred to the water by the fan. We can use the formula:

Q = P * t

Where: - Q is the heat transferred to the water (in joules) - P is the power of the motor (in watts) - t is the time for which the motor operates (in seconds)

In this case, we are given that the power of the motor is 50 W and the time for which it operates is 10 minutes. We need to convert the time to seconds before using it in the formula.

t = 10 minutes = 10 * 60 seconds = 600 seconds

Now we can calculate the heat transferred to the water:

Q = 50 W * 600 s = 30000 J

The heat transferred to the water is 30000 joules.

Next, we need to calculate the change in temperature of the water using the formula:

Q = m * c * ΔT

Where: - Q is the heat transferred to the water (in joules) - m is the mass of the water (in kilograms) - c is the specific heat capacity of water (in joules per kilogram per degree Celsius) - ΔT is the change in temperature of the water (in degrees Celsius)

In this case, we are given that the mass of the water is 2 kg. The specific heat capacity of water is approximately 4186 J/kg°C.

Substituting the values into the formula, we can solve for ΔT:

30000 J = 2 kg * 4186 J/kg°C * ΔT

Simplifying the equation:

ΔT = 30000 J / (2 kg * 4186 J/kg°C) ≈ 3.59°C

Therefore, the temperature of the water will increase by approximately 3.59 degrees Celsius.

Answer

The temperature of the water inside the calorimeter will increase by approximately 3.59 degrees Celsius after the fan, powered by a 50 W motor, rotates the blades for 10 minutes.