Question

Двигатель мощностью 52 Вт в течении 10 минут вращает лопасти вентилятора внутри калориметра с водой. На сколько повысится за это время температура воды, если её масса 2 кг.?

Answer 1

Q=52 Вт *10 минут=31200 Дж

дельта Т=Q/cm=31200Дж/4200Дж/кгК*2кг=3.71 град

Answer 2

10 минут = 600 секунд

52Вата втечение 600 секунд 

Q (теплота) = 52*600 = 31200 Джоулей.

Чтобы нагреть 1 кг воды на 1 градус, необходимо 4200 Джоулей.

Найдем нергию необходимую для нагрева 2кг на 1 градус

Qнеобх = 4200*2 = 8400 Дж/градус

У нас 31200 джоулей,

T разница = 31200/8400 примерно равно 3.71 градуса.

 

 

Answer 3

Calculation of Temperature Increase

To calculate the increase in temperature of the water inside the calorimeter, we can use the formula:

Q = mcΔT

Where: - Q is the heat energy transferred to the water (in joules) - m is the mass of the water (in kilograms) - c is the specific heat capacity of water (approximately 4.18 J/g°C or 4180 J/kg°C) - ΔT is the change in temperature of the water (in degrees Celsius)

In this case, the mass of the water is given as 2 kg, and we need to find the change in temperature (ΔT) when a 52 W engine rotates the fan blades inside the calorimeter for 10 minutes.

Conversion of Power to Energy

First, we need to convert the power of the engine from watts to joules per second (J/s) by multiplying it by the time in seconds:

Energy = Power × Time

Given: - Power = 52 W - Time = 10 minutes = 600 seconds

Energy = 52 W × 600 s = 31,200 J

Calculation of Temperature Increase

Now that we have the energy transferred to the water, we can calculate the change in temperature using the formula mentioned earlier:

Q = mcΔT

Substituting the known values: - Q = 31,200 J - m = 2 kg - c = 4180 J/kg°C

31,200 J = (2 kg) × (4180 J/kg°C) × ΔT

Simplifying the equation:

ΔT = 31,200 J / (2 kg × 4180 J/kg°C)

ΔT ≈ 3.73°C

Therefore, the temperature of the water inside the calorimeter will increase by approximately 3.73 degrees Celsius after the 52 W engine rotates the fan blades for 10 minutes.

Please note that this calculation assumes no heat loss to the surroundings and that the specific heat capacity of water remains constant throughout the process.