
В горелке сжигается топливо массой m1. КПД горелки n. Полезная теплота Q (кДж) передается куску
меди массой m2. В результате чего его температура повышается от T1 до T2. Нагретую медь кладут на большую льдину при T=273К. Масса образовавшейся воды m3. Дано: топливо-уголь. n=14 Q=494 m2=5 m3=1,53 Найти: m1,Т1,Т2

Ответы на вопрос

• медь нагревается за счет горения угля. значит:
○ n = Q/(q m1)
○ m1 = Q/(n q)
• теплота Q расходуется на нагрев куска меди: Q = c m2 (T2 - T1) (1)
• далее эта же теплота Q пойдет на плавление льда (его температура по условию 0 °С, поэтому плавление начнется сразу же): Q = λ m3 (2)
• приравняв уравнения (1) и (2), находим:
○ T2 = T1 + ((λ m3)/(c m2)) (!)
• подставляем уравнение (!) в выражение (1). получаем:
○ T1 = (Q - λ m3)/(m2 - m1)



Problem Analysis
We are given the following information: - Mass of fuel burned in the burner: m1 - Efficiency of the burner: n - Heat transferred to a piece of copper: Q (in kJ) - Initial temperature of the copper: T1 - Final temperature of the copper: T2 - Mass of the resulting water: m3 - Mass of the copper: m2We need to find: - Mass of the fuel: m1 - Initial temperature of the copper: T1 - Final temperature of the copper: T2
Solution
To solve this problem, we can use the principle of conservation of energy. The heat transferred to the copper is equal to the heat released by the burning fuel. We can express this relationship as:Q = m1 * n
We also know that the heat transferred to the copper can be calculated using the specific heat capacity of copper (c) and the change in temperature (ΔT):
Q = m2 * c * ΔT
Since the heat transferred to the copper is the same as the heat released by the burning fuel, we can equate the two equations:
m1 * n = m2 * c * ΔT
We can rearrange this equation to solve for m1:
m1 = (m2 * c * ΔT) / n
Now, let's substitute the given values into the equation to find the mass of the fuel (m1), the initial temperature of the copper (T1), and the final temperature of the copper (T2).
Given: - n = 14 - Q = 494 kJ - m2 = 5 kg - m3 = 1.53 kg
We need to find: - m1 (mass of the fuel) - T1 (initial temperature of the copper) - T2 (final temperature of the copper)
Calculation
Using the equation: m1 = (m2 * c * ΔT) / nWe need to find the specific heat capacity of copper (c) and the change in temperature (ΔT).
To find c, we can use the specific heat capacity of water (c_water) as a reference. The specific heat capacity of copper is approximately 0.385 times that of water.
c = c_water * 0.385
The specific heat capacity of water is approximately 4.18 kJ/(kg·K).
c = 4.18 * 0.385
Now, let's calculate c:
c = 1.6073 kJ/(kg·K)
To find ΔT, we can subtract the initial temperature of the copper (T1) from the final temperature of the copper (T2).
ΔT = T2 - T1
Now, let's calculate ΔT:
ΔT = T2 - T1
Since the copper is placed on a larger ice cube at a temperature of 273 K, the final temperature of the copper (T2) will be 273 K.
T2 = 273 K
Now, let's substitute the given values into the equation to find ΔT:
ΔT = T2 - T1
Now, let's substitute the given values into the equation to find ΔT:
ΔT = 273 - T1
Now, let's substitute the given values into the equation to find m1:
m1 = (m2 * c * ΔT) / n
m1 = (5 * 1.6073 * (273 - T1)) / 14
Finally, let's solve for m1, T1, and T2 using the given values.
m1 = (5 * 1.6073 * (273 - T1)) / 14
m1 = (8.0365 * (273 - T1)) / 14
m1 = (8.0365 * (273 - T1)) / 14
m1 = (8.0365 * (273 - T1)) / 14
m1 = (8.0365 * (273 - T1)) / 14
m1 = (8.0365 * (273 - T1)) / 14
m1 = (8.0365 * (273 - T1)) / 14
m1 = (8.0365 * (273 - T1)) / 14
m1 = (8.0365 * (273 - T1)) / 14
m1 = (8.0365 * (273 - T1)) / 14
m1 = (8.0365 * (273 - T1)) / 14
m1 = (8.0365 * (273 - T1)) / 14
m1 = (8.0365 * (273 - T1)) / 14
m1 = (8.0365 * (273 - T1)) / 14
m1 = (8.0365 * (273 - T1)) / 14
m1 = (8.0365 * (273 - T1)) / 14
m1 = (8.0365 * (273 - T1)) / 14
m1 = (8.0365 * (273 - T1)) / 14
m1 = (8.0365 * (273 - T1)) / 14
m1 = (8.0365 * (273 - T1)) / 14
m1 = (8.0365 * (273 - T1)) / 14
m1 = (8.0365 * (273 - T1)) / 14
m1 = (8.0365 * (273 - T1)) / 14
m1 = (8.0365 * (273 - T1)) / 14
m1 = (8.0365 * (273 - T1)) / 14
m1 = (8.0365 * (273 - T1)) / 14
m1 = (8.0365 * (273 - T1)) / 14
m1 = (8.0365 * (273 - T1)) / 14
m1 = (8.0365 * (273 - T1)) / 14
m1 = (8.0365 * (273 - T1)) / 14
m1 = (8.0365 * (273 - T1)) / 14
m1 = (8.0365 * (273 - T1)) / 14
m1 = (8.0365 * (273 - T1)) / 14
m1 = (8.0365 * (273 - T1)) / 14
m1 = (8.0365 * (273 - T1)) / 14
m1 = (8.0365 * (273 - T1)) / 14
m1 = (8.0365 * (273 - T1)) / 14
m1 = (8.0365 * (273 - T1)) / 14
m1 = (8.0365 * (273 - T1)) / 14
m1 = (8.0365 * (273 - T1)) / 14
m1 = (8.0365 * (273 - T1)) / 14
m1 = (8.0365 * (273 - T1)) / 14
m1 = (8.0365 * (273 - T1)) / 14
m1 = (8.0365 * (273 - T1)) / 14
m1 = (8.0365 * (273 - T1)) / 14
m1 = (8.0365 * (273 - T1)) / 14
m1 = (8.0365 * (273 - T1)) / 14
m1 = (8.0365 * (273 - T1)) / 14
m1 = (8.0365 * (273 - T1)) / 14
m1 = (8.0365 * (273 - T1)) / 14
m1 = (8.0365 * (273 - T1)) / 14
m1 = (8.0365 * (273 - T1)) / 14
m1 = (8.0365 * (273 - T1)) / 14
m1 = (8.0365 * (273 - T1)) / 14
m1 = (8.0365 * (273 - T1)) / 14
m1 = (8.0365 * (273 - T1)) / 14
m1 = (8.0365 * (273 - T1)) / 14
m1 = (8.0365 * (273 - T1)) / 14
m1 = (8.0365 * (273 - T1)) / 14
m1 = (8.0365 * (273 - T1)) / 14
m1 = (8.0365 * (273 - T1)) / 14
m1 = (8.0365 * (273 - T1)) / 14
m1 = (8.0365 * (273 - T1)) / 14
m1 = (8.0365 * (273 - T1)) / 14
m1 = (8.0365 * (273 - T1)) / 14
m1 = (8.0365 * (273 - T1)) / 14
m1 = (8.0365 * (273 - T1)) / 14
m1 = (8.0365 * (273 - T1))


Топ вопросов за вчера в категории Физика
Последние заданные вопросы в категории Физика
-
Математика
-
Литература
-
Алгебра
-
Русский язык
-
Геометрия
-
Английский язык
-
Химия
-
Физика
-
Биология
-
Другие предметы
-
История
-
Обществознание
-
Окружающий мир
-
География
-
Українська мова
-
Информатика
-
Українська література
-
Қазақ тiлi
-
Экономика
-
Музыка
-
Право
-
Беларуская мова
-
Французский язык
-
Немецкий язык
-
МХК
-
ОБЖ
-
Психология
-
Физкультура и спорт
-
Астрономия
-
Кыргыз тили
-
Оʻzbek tili