Question

Алюминиевый сосуд массой 250 г наполнен 3,5 л воды.Какое количество теплоты должно быть передано ему от пламени газовой плиты для увеличения температуры воды с 20°C до 60°C ?

Answer 1

Удельная теплота алюминия 920 Дж/кг*С
удельная теплота воды 4200 Дж/кг*С
Q=cm(t1-t2)
920*0/25*(60-20)=9200 Дж = 9.2 кДж - алюминий
4200*3.5*(60-20)=588000 Дж = 588 кДж - вода
9.2+588= 597.2 кДж

Answer 2

Calculating the Heat Transfer for Heating Water in an Aluminum Container

To calculate the amount of heat required to increase the temperature of the water in the aluminum container, we can use the formula:

Q = mcΔT

Where: - Q = heat energy (in joules) - m = mass of the water (in kg) - c = specific heat capacity of water (in J/g°C or J/kg°C) - ΔT = change in temperature (in °C)

Given: - Mass of the aluminum container, m_aluminum = 250 g = 0.25 kg - Mass of the water, m_water = 3.5 L = 3.5 kg - Initial temperature, T_initial = 20°C - Final temperature, T_final = 60°C

Specific Heat Capacity of Water

The specific heat capacity of water is approximately 4.18 J/g°C or 4186 J/kg°C.

Calculating the Heat Transfer

Using the formula Q = mcΔT, we can calculate the heat transfer required to increase the temperature of the water in the aluminum container from 20°C to 60°C.

First, we need to calculate the mass of the water and the aluminum container combined: m_total = m_aluminum + m_water

Then, we can calculate the heat transfer: Q = m_total * c * ΔT

Substitute the values: Q = (0.25 kg + 3.5 kg) * 4186 J/kg°C * (60°C - 20°C)

Calculating the value: Q ≈ 3.75 * 4186 * 40

Q ≈ 627,900 J

So, the amount of heat that must be transferred from the flame of the gas stove to the aluminum container to increase the temperature of the water from 20°C to 60°C is approximately 627,900 joules.

Therefore, approximately 627,900 joules of heat must be transferred to the aluminum container from the gas stove's flame.

I hope this helps! If you have further questions, feel free to ask.