Question

1)Какое количество теплоты отдаст стакан кипятка (250 мл), остывая до температуры 14 ͦ С 2)Какое количество теплоты получили алюминиевая кастрюля массой 200г и находящаяся в ней вода объемом 1,5 л при нагревании от 20 ͦ С до кипения? 3)Когда в бак с водой добавили еще 3 л кипятка и перемешали всю воду, то температура воды в баке стала равна 35 ͦ С. Пренебрегая потерями теплоты на нагревание бака и окружающей среды, определите начальный объем воды в баке.

Answer 1

1) V=250 мл = 0,00025 куб м
p=1000 кг/куб м
с=4200 Дж/кг*град С
t1=100 град С
t2=16 град С
m=1000*0,00025=0,25 кг
Q=0,25*4200*(100-16)=88200 Дж
2)с воды=4200 Дж/кг*град С
m кастр =200 г = 0,2 кг
V=1,5 л=0,0015 куб м
p=1000 кг/куб м
с алюм = 920 Дж/кг*град С
t1=20
t2=100
Найти Q общий
Решение
m=0,0015*1000=0,15 кг
Q=4200*0,15*(100-20)+920*0,2*(100-20)=65120 Дж
Ответ: 65120 Дж
3)х+3*100=35(х+3)
х+300=35х+105
х-35х=-300+105
-34х=-195
34х=195
х=5,7 л
Ответ: 5,7 л ~ 6 л

Answer 2

Calculation of Heat Transfer

1) To calculate the amount of heat released by a glass of boiling water (250 ml) as it cools down to a temperature of 14°C, we can use the formula:

Q = mcΔT

Where: - Q is the heat transfer (in joules) - m is the mass of the substance (in grams) - c is the specific heat capacity of the substance (in J/g°C) - ΔT is the change in temperature (in °C)

In this case, the substance is water, which has a specific heat capacity of approximately 4.18 J/g°C.

First, we need to convert the volume of water (250 ml) to grams. Since the density of water is approximately 1 g/ml, the mass of the water is 250 grams.

Now we can calculate the heat transfer:

Q = (250 g) * (4.18 J/g°C) * (100°C - 14°C)

Calculating this expression gives us the amount of heat transferred by the glass of boiling water as it cools down to 14°C.

2) To calculate the amount of heat received by an aluminum pot (mass 200 g) containing 1.5 liters of water as it is heated from 20°C to boiling point, we can use the same formula:

Q = mcΔT

The specific heat capacity of aluminum is approximately 0.897 J/g°C.

First, we need to convert the volume of water (1.5 liters) to grams. Since the density of water is approximately 1 g/ml, the mass of the water is 1500 grams.

Now we can calculate the heat transfer:

Q = (200 g + 1500 g) * (0.897 J/g°C) * (100°C - 20°C)

Calculating this expression gives us the amount of heat transferred to the aluminum pot and the water as it is heated from 20°C to the boiling point.

3) To determine the initial volume of water in the tank when 3 liters of boiling water are added and the temperature of the water in the tank becomes 35°C, we can assume that the heat transfer is negligible. This means that the heat gained by the water is equal to the heat lost by the boiling water.

Using the formula Q = mcΔT, we can set up the equation:

Q1 = Q2

Where: - Q1 is the heat gained by the water in the tank - Q2 is the heat lost by the boiling water

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

Let's assume the initial volume of water in the tank is V liters. The mass of the water in the tank is then 1000V grams.

For the boiling water, the mass is 3000 grams (3 liters * 1000 grams/liter).

Setting up the equation:

(1000V g) * (4.18 J/g°C) * (35°C - 20°C) = (3000 g) * (4.18 J/g°C) * (100°C - 35°C)

Simplifying and solving for V will give us the initial volume of water in the tank.

Please note that these calculations assume ideal conditions and neglect any heat losses to the surroundings or the container itself.