Question

В алюминиевый калориметр массой 0,2 кг , содержащий 0,3 кг воды , при 23 градусах Цельсия опустили 80 грамм льда при 0 градусах Цельсия. Какая установится температура после таяния льда?

Answer 1

Объяснение:

Дано:

c(а)=920Дж/кг⁰С

с(вода)=4200Дж/кг⁰С

/l=34×10⁴Дж/кг

m(к)=0,2кг

m(в)=0,3кг

t(лед)=0⁰С

m(л)=0,08кг

t(в и к)=23⁰С

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Найти установившиюся температуру(t)-?

Нужные формулы:

Q=c×m×(t1-t); Q=/|×m; Нужно найти здесь t.

Расчитываем:

Q=Q(вода)+Q(калориметр)

(Аляюминия и воду, теплоту Q)

Q(в)=cmt=920×0,2×(23-t)=4232-184t

Q(к)=cmt=4200×0,3×(23-t)=28980-1260t

Q=33212-1444t

Q--- мы нашли осталось Плавления льда

и снова теплота для воды.

Q= /l × m Где---/l это её плавление

Q=340000×0,08=27200Дж

Q= 4200×0,08×t

Зачем умножаем на 4200 а не на

теплоёмкость лёда?

Умножаем потому что, через воду лёд

получает тепло.

Q=336t

Теперь мы приравниваем Найденные нами значения что-бы найти t

33212-1444t=27200+336t

33212-27200=336t+1444t

6012Дж=1780t

t=6012/1780=~3,4⁰C будет вода после

таяния льда.

Answer 2

Calculation of Final Temperature after Ice Melting

To determine the final temperature after the ice melts in the aluminum calorimeter, we can use the principle of conservation of energy. The heat gained by the ice as it melts will be equal to the heat lost by the water and the calorimeter.

The heat gained by the ice can be calculated using the formula:

Q_ice = m_ice * L_f

Where: - Q_ice is the heat gained by the ice (in Joules) - m_ice is the mass of the ice (in kilograms) - L_f is the latent heat of fusion for water, which is 334,000 J/kg

In this case, the mass of the ice is 80 grams, which is equal to 0.08 kilograms. Therefore:

Q_ice = 0.08 kg * 334,000 J/kg = 26,720 J

The heat lost by the water and the calorimeter can be calculated using the formula:

Q_water_calorimeter = (m_water + m_calorimeter) * c * ΔT

Where: - Q_water_calorimeter is the heat lost by the water and the calorimeter (in Joules) - m_water is the mass of the water (in kilograms) - m_calorimeter is the mass of the calorimeter (in kilograms) - c is the specific heat capacity of water, which is approximately 4,186 J/kg°C - ΔT is the change in temperature (in °C)

In this case, the mass of the water is 0.3 kilograms, the mass of the calorimeter is 0.2 kilograms, and the initial temperature is 23°C. The final temperature after the ice melts is denoted by T_f. Therefore:

Q_water_calorimeter = (0.3 kg + 0.2 kg) * 4,186 J/kg°C * (T_f - 23°C)

Since the heat gained by the ice is equal to the heat lost by the water and the calorimeter, we can set up the equation:

Q_ice = Q_water_calorimeter

26,720 J = (0.5 kg) * 4,186 J/kg°C * (T_f - 23°C)

Simplifying the equation:

26,720 J = 2,093 J/°C * (T_f - 23°C)

Dividing both sides of the equation by 2,093 J/°C:

12.77°C = T_f - 23°C

Adding 23°C to both sides of the equation:

T_f = 12.77°C + 23°C = 35.77°C

Therefore, the final temperature after the ice melts in the aluminum calorimeter will be approximately 35.77°C.