Question

На примусе в медной кастрюле массой 2 кг нагревается от 10 с прим На до кипения 5 кг воды. Кид примуса 40%. Определите, какое количество керосина для этого затратили. (Примус бесеbumuльный наeреваmeльный прибор, работающий на бензине или керосине)

Answer 1

1) Нагреваем воду до кипения:
Q1 =c1*m1*(t-t1) = 4200*5*(100-10) = 3,78*10⁶ Дж
2) Нагреваем кастрюлю:
Q2=c2*m2*(t-t1) = 385*2*(100-10) = 0,07*10⁶ Дж

3)Суммарное количество теплоты:
Q = Q1+Q2 = 3,85*10⁶ Дж    

4) Сжигаем керосин:
Q₃ = q*m = 46,2*10⁶ *m

КПД = Q / Q₃
0,40 = 3,85*10⁶ / (46,2*10⁶*m)
0,40 = 0,083 / m
m = 0,083 / 0,40 ≈ 0,210 кг   или   210 грамм

Answer 2

Calculation of the Amount of Kerosene Used

To calculate the amount of kerosene used in heating the 5 kg of water from 10°C to boiling point, we need to consider the heat transfer and energy requirements involved.

The heat transfer equation can be used to calculate the amount of heat required to raise the temperature of the water:

Q = m * c * ΔT

Where: - Q is the heat energy required (in Joules) - m is the mass of the water (in kg) - c is the specific heat capacity of water (approximately 4.18 J/g°C) - ΔT is the change in temperature (in °C)

First, let's convert the mass of water from 5 kg to grams:

m = 5,000 g

Next, let's calculate the change in temperature:

ΔT = 100°C - 10°C = 90°C

Now, we can calculate the heat energy required:

Q = 5,000 g * 4.18 J/g°C * 90°C = 1,881,000 J

Since the efficiency of the stove is given as 40%, we need to account for the energy loss. The amount of kerosene used will be 40% of the total energy required:

Amount of kerosene used = 0.4 * 1,881,000 J = 752,400 J

Please note that this calculation assumes ideal conditions and does not take into account factors such as heat loss to the surroundings or the efficiency of the stove. It provides an estimate based on the given information.