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3 years ago

If heat energy is continually being added from the hotplate why doesn't the temperature of the water increase at a steady rate?

Chemistry

2 answers:

Dovator [93]3 years ago

8 0

Hello! Allow me to help!

Your question: If heat energy is continually being added from the hotplate why doesn't the temperature of the water increase at a steady rate?

My answer: Because at boiling point liquid getting converted into gaseous state, and now all given temperature is utilized in conversion of liquid to gaseous state, so no change in temperature occurs.

Why is my answer correct? Good question! Allow me to explain: Because adding heat energy usually results in a temperature rise, people often confuse heat and temperature. In common speech, the two terms mean the same: "I will heat it" means you will add heat; "I will warm it up" means you will increase the temperature. No one usually bothers to distinguish between these. Adding heat, however, does not always increase the temperature. For instance, when water is boiling, adding heat does not increase its temperature. This happens at the boiling temperature of every substance that can vaporize. At the boiling temperature, adding heat energy converts the liquid into a gas WITHOUT RAISING THE TEMPERATURE. Adding heat to a boiling liquid is an important exception to general rule that more heat makes a higher temperature. When energy is added to a liquid at the boiling temperature, its converts the liquid into a gas at the same temperature. In this case, the energy added to the liquid goes into breaking the bonds between the liquid molecules without causing the temperature to change. The same thing happens when a solid changes into liquid. For instance, ice and water can exist together at the melting temperature. Adding heat to an ice-water slush will convert some of the ice to water without changing the temperature. In general, whenever there is a change of state, such as the solid-liquid or the liquid-gas transition, heat energy can be added without a temperature change. The change of state requires energy, so added energy goes into that instead of increasing the temperature.

Hope this helps! UwU

-Maxwell

Sonbull [250]3 years ago

6 0

Sheesh that person wrote a lot but I agree with them!

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At which temperature do the molecules of an ideal gas have 3 times the kinetic energy they have at 32of?

algol [13]

Answer:

820 K

Explanation:

As per Boltzman equation, kinetic energy (KE) is in direct relation to the temperature, measured in absolute scale Kelvin.

KE α T.

Then, the temperature at which the molecules of an ideal gas have 3 times the kinetic energy they have at any given temperature will be 3 times such temperature.

So, you must just convert the given temperature, 32°F, to kelvin scale.

You can do that in two stages.

First, convert 32°F to °C. Since, 32°F is the freezing temperature of water, you may remember that is 0°C. You can also use the conversion formula: T (°C) = [T (°F) - 32] / 1.80

Second, convert 0°C to kelvin:

T (K) = T(°C) + 273.15 K= 273.15 K

Then, 3 times gives you: 3 × 273.15 K = 819.45 K

Since, 32°F has two significant figures, you must report your answer with the same number of significan figures. That is 820 K.

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3 years ago

The solubility of oxygen in lakes high in the Rocky Mountains is affected by the altitude. If the solubility of O2 from the air

IceJOKER [234]

Answer:

$1.75\cdot 10^{-4} M$

Explanation:

Henry's law states that the solubility of a gas is directly proportional to its partial pressure. The equation may be written as:

$S = k_H p^o$

Where $k_H$ is Henry's law constant.

Our strategy will be to identify the Henry's law constant for oxygen given the initial conditions and then use it to find the solubility at different conditions.

Given initially:

$S_1 = 2.67\cdot 10^{-4} M$

Also, at sea level, we have an atmospheric pressure of:

$p = 1.00 atm$

Given mole fraction:

$\chi_{O_2} = 0.209$

According to Dalton's law of partial pressures, the partial pressure of oxygen is equal to the product of its mole fraction and the total pressure:

$p^o = \chi_{O_2} p$

Then the equation becomes:

$S_1 = k_H \chi_{O_2} p$

Solve for $k_H$ :

$k_H = \frac{S_1}{\chi_{O_2} p} = \frac{2.67\cdot 10^{-4} M}{0.209\cdot 1.00 atm} = 0.001278 M/atm$

Now we're given that at an altitude of 12,000 ft, the atmospheric pressure is now:

$p = 0.657 atm$

Apply Henry's law using the constant we found:

$S_2 = k_H \chi_{O_2} p = 0.001278 M/atm\cdot 0.209\cdot 0.657 atm = 1.75\cdot 10^{-4} M$

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bezimeni [28]

Answer:

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Rashid [163]

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