Question

Какое количество теплоты необходимо для нагревания 3 л воды в алюминиевой кастрюле массой 300 г от 20 до 100 градусов Цельсия? (Удельная теплоемкость воды 4200 Дж/(кг С), алюминия 920 Дж/(кг С) , плотность воды 1000 кг/м3)

Answer 1

Q = Q1 + Q2
Q = c1 m1 (t - t1) + c2 m2 (t - t1)
Q = (t - t1) (c1 V1ρ1 + c2 m2)
Q = (100 - 20) (4200 * 0.003 * 1000 + 920 *  0.3)
Q = 1030000 Дж
Ответ: 1030 кДж

Answer 2

Calculating Heat Required to Heat Water in an Aluminum Pot

To calculate the amount of heat required to heat 3 liters of water in an aluminum pot from 20 to 100 degrees Celsius, we can use the formula:

Q = mcΔT

Where: - Q = heat energy (in joules) - m = mass of the substance (in kilograms) - c = specific heat capacity of the substance (in joules per kilogram per degree Celsius) - ΔT = change in temperature (in degrees Celsius)

Given data: - Mass of water, m = 3 kg (since 1 liter of water has a mass of 1 kg) - Specific heat capacity of water, c_water = 4200 J/(kg°C) - Change in temperature, ΔT = 100°C - 20°C = 80°C

Let's calculate the heat required step by step.

First, we calculate the heat required to heat the water: Q_water = m_water * c_water * ΔT_water

Q_water = 3 kg * 4200 J/(kg°C) * 80°C Q_water = 1008000 J

Now, we calculate the heat required to heat the aluminum pot: The specific heat capacity of aluminum, c_aluminum = 920 J/(kg°C)

Q_aluminum = m_aluminum * c_aluminum * ΔT_aluminum

The mass of the aluminum pot, m_aluminum = 0.3 kg (since the mass is given in grams and needs to be converted to kilograms) ΔT_aluminum = 100°C - 20°C = 80°C

Q_aluminum = 0.3 kg * 920 J/(kg°C) * 80°C Q_aluminum = 22080 J

Finally, we add the heat required for both the water and the aluminum pot: Total heat required = Q_water + Q_aluminum Total heat required = 1008000 J + 22080 J Total heat required = 1030080 J

So, the total amount of heat required to heat 3 liters of water in an aluminum pot with a mass of 300 g from 20 to 100 degrees Celsius is 1030080 joules.