Question

воду взятую при температуре кипения массой 100г полностью превратили в пар. найти кол-во теплоты необходимое для парообразования. ответ в кДж​

Answer 1

Ответ:

230кДж

Объяснение:

Дано:

m=100г=0,1кг

L=2,3×10⁶Дж/кг

Решение:Q=Lm=2,3×10⁶Дж/кг×0,1кг=0,23×10⁶Дж

0,23×10⁶Дж=230000Дж=230кДж

Answer 2

Calculation of Heat Required for Vaporization

To find the amount of heat required to completely vaporize 100g of water at its boiling point, we can use the formula:

Q = m * ΔHvap

Where: - Q is the heat required for vaporization - m is the mass of the substance (in this case, 100g) - ΔHvap is the heat of vaporization of water

The heat of vaporization of water is the amount of heat energy required to convert 1 gram of water from its liquid state to its gaseous state at its boiling point. The heat of vaporization of water is approximately 40.7 kJ/mol.

To calculate the heat required for vaporization, we need to convert the mass of water from grams to moles. The molar mass of water (H2O) is approximately 18.015 g/mol.

Let's calculate the number of moles of water:

moles = mass / molar mass

moles = 100g / 18.015 g/mol

Using this value, we can calculate the heat required for vaporization:

Q = moles * ΔHvap

Now, let's substitute the values into the equation:

Q = (100g / 18.015 g/mol) * 40.7 kJ/mol

Calculating this expression, we find that the heat required for vaporization of 100g of water is approximately 226.1 kJ.

Please note that the calculated value is an approximation and may vary slightly depending on the specific heat of vaporization used.

I hope this helps! Let me know if you have any further questions.