Question

Отдаю 15 пкт + 8 за лучшее решениеВ холодильнике изготовили 500 см3 льда при температуре -5 градусов цельсия. Сколько теплоты было отведено при этом от воды и льда, если начальная тепмература воды была 15 градусов цельсия

Answer 1

1.удельная теплота воды*объем*плотность*15

2.плавление(лямде)*объем*плотность

3. удельная теплота льда*объем*плотность*5

4. все сложить

 

сорь, таблички с данными под рукой нет

Answer 2

Calculation of Heat Transferred from Water and Ice

To calculate the amount of heat transferred from water and ice, we can use the equation:

Q = m * c * ΔT

Where: - Q is the heat transferred - m is the mass of the substance - c is the specific heat capacity of the substance - ΔT is the change in temperature

In this case, we need to calculate the heat transferred from the water and ice when the initial temperature of the water was 15 degrees Celsius and 500 cm^3 of ice was formed at a temperature of -5 degrees Celsius.

Calculation for Water:

The specific heat capacity of water is approximately 4.18 J/g°C.

First, we need to calculate the mass of the water. We know that the volume of the water is 500 cm^3, but we need to convert it to grams. Since the density of water is 1 g/cm^3, the mass of the water is also 500 grams.

Next, we can calculate the change in temperature (ΔT) for the water. The initial temperature of the water was 15 degrees Celsius, and since it turned into ice, the final temperature is 0 degrees Celsius. Therefore, ΔT = 0 - 15 = -15 degrees Celsius.

Now we can calculate the heat transferred from the water:

Q_water = m_water * c_water * ΔT_water

Q_water = 500 g * 4.18 J/g°C * (-15°C)

Calculating this, we get:

Q_water = -31,350 J

Therefore, the heat transferred from the water is approximately -31,350 Joules.

Calculation for Ice:

The specific heat capacity of ice is approximately 2.09 J/g°C.

The mass of the ice is the same as the mass of the water, which is 500 grams.

The change in temperature (ΔT) for the ice is the final temperature of the ice (0 degrees Celsius) minus the initial temperature of the ice (-5 degrees Celsius). Therefore, ΔT = 0 - (-5) = 5 degrees Celsius.

Now we can calculate the heat transferred from the ice:

Q_ice = m_ice * c_ice * ΔT_ice

Q_ice = 500 g * 2.09 J/g°C * 5°C

Calculating this, we get:

Q_ice = 5,225 J

Therefore, the heat transferred from the ice is approximately 5,225 Joules.

Total Heat Transferred:

To find the total heat transferred from the water and ice, we can simply add the heat transferred from each component:

Total Heat Transferred = Q_water + Q_ice

Total Heat Transferred = -31,350 J + 5,225 J

Calculating this, we get:

Total Heat Transferred = -26,125 J

Therefore, the total heat transferred from the water and ice is approximately -26,125 Joules.

Please note that the negative sign indicates that heat was transferred out of the system (water and ice) into the surroundings.