Question

В алюминиевую кастрюлю массой 1400 грамм при нормальных условиях положили лёд при температуре -4°С, после его расстаянии до 5°С получили 1 литр воды и довели её до кипения. Какое количество теплоты получили или отдали вода и кастрюля по отдельности?

Answer 1

Надеюсь будет правильно
найдём Q кострюли
Q=m*c*∆t=1,4*920*104=133952дж
теперь найдём Q воды
Q=m*c*∆t=v*плотность*с*∆t=0.001*1000*4200*(100-5)=399000дж
теперь найдём Q льда
Q=1кг(тк масса воды равна 1 кг)*2100*(5+4)=18900дж
теперь (возможно) надо суммировать Q воды и льда: 399000+18900=417900
ответ: Q воды=417900, Q кастрюли = 133952

Answer 2

Calculation of Heat Transfer

To calculate the amount of heat transferred by the water and the aluminum pot, we need to consider the specific heat capacities of both substances and the temperature change they undergo.

The specific heat capacity of water is approximately 4.18 J/g°C This means that it takes 4.18 Joules of energy to raise the temperature of 1 gram of water by 1 degree Celsius.

The specific heat capacity of aluminum is approximately 0.897 J/g°C This means that it takes 0.897 Joules of energy to raise the temperature of 1 gram of aluminum by 1 degree Celsius.

Now, let's calculate the heat transfer for each component separately.

Heat Transfer for the Water

Given: - Mass of water = 1 liter = 1000 grams - Initial temperature of water = 5°C - Final temperature of water = 100°C (boiling point)

The temperature change for the water is: ΔT = Final temperature - Initial temperature = 100°C - 5°C = 95°C

The heat transfer for the water can be calculated using the formula: Q = m * c * ΔT

Where: - Q is the heat transfer - m is the mass of the water - c is the specific heat capacity of water - ΔT is the temperature change

Substituting the values: Q = 1000 g * 4.18 J/g°C * 95°C = 395,900 Joules

Therefore, the water received or absorbed 395,900 Joules of heat.

Heat Transfer for the Aluminum Pot

Given: - Mass of the aluminum pot = 1400 grams - Initial temperature of the aluminum pot = -4°C - Final temperature of the aluminum pot = 100°C (boiling point)

The temperature change for the aluminum pot is: ΔT = Final temperature - Initial temperature = 100°C - (-4°C) = 104°C

The heat transfer for the aluminum pot can be calculated using the same formula: Q = m * c * ΔT

Where: - Q is the heat transfer - m is the mass of the aluminum pot - c is the specific heat capacity of aluminum - ΔT is the temperature change

Substituting the values: Q = 1400 g * 0.897 J/g°C * 104°C = 130,643.2 Joules

Therefore, the aluminum pot received or absorbed 130,643.2 Joules of heat.

Conclusion

In summary, the water received or absorbed 395,900 Joules of heat, while the aluminum pot received or absorbed 130,643.2 Joules of heat.

Please note that these calculations assume that there is no heat loss to the surroundings during the process.