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4 years ago

Heat is transferred from the heating element to the pot to the boiling water. What is the MOST likely method of heat transfer?

Chemistry

2 answers:

madreJ [45]4 years ago

7 0

Answer: Conduction

Explanation: There are three modes of heat transfer.

1. Conduction: This type of heat transfer happens when there is direct contact between the two object.

2. Convection: This type of heat transfer happens when there is a movement of fluid (liquid or gas).

3. Radiation: This type of heat transfer happens when there is direct transfer of energy through space.

Thus heat being transferred from the heating element to the pot to the boiling water is an example of conduction as the they are in direct contact with each other.

Whitepunk [10]4 years ago

5 0

The answer is CONDUCTION :D

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A sample of a substance has a mass of 4.2 grams and a volume of 6 milliliters. The density of this substance is

Tanzania [10]

Hello!

Mass =4.2 g

Volume =6 mL

Therefore:

Density = mass / volume

Density = 4.2 / 6

Density = 0.7 g/mL

Hope that helps!

4 0

3 years ago

Read 2 more answers

What are 3 things an Apple, Knife and Mirror have in common?

s2008m [1.1K]

1)They are all take up space.
2)They all have mass.
3)They are all solids.

Because of 1 and 2, they are all matter.

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3 years ago

Which statements accurately describe volume? Check all that apply.

lesantik [10]

Where is the chooses for you question

8 0

3 years ago

The normal boiling point of bromine is 58.8°C, and its enthalpy of vaporization is 30.91 kJ/mol. What is the approximate vapor p

saul85 [17]

Answer : The vapor pressure of bromine at $10.0^oC$ is 0.1448 atm.

Explanation :

The Clausius- Clapeyron equation is :

$\ln (\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}\times (\frac{1}{T_1}-\frac{1}{T_2})$

where,

$P_1$ = vapor pressure of bromine at $10.0^oC$ = ?

$P_2$ = vapor pressure of propane at normal boiling point = 1 atm

$T_1$ = temperature of propane = $10.0^oC=273+10.0=283.0K$

$T_2$ = normal boiling point of bromine = $58.8^oC=273+58.8=331.8K$

$\Delta H_{vap}$ = heat of vaporization = 30.91 kJ/mole = 30910 J/mole

R = universal constant = 8.314 J/K.mole

Now put all the given values in the above formula, we get:

$\ln (\frac{1atm}{P_1})=\frac{30910J/mole}{8.314J/K.mole}\times (\frac{1}{283.0K}-\frac{1}{331.8K})$

$P_1=0.1448atm$

Hence, the vapor pressure of bromine at $10.0^oC$ is 0.1448 atm.

4 0

3 years ago

A container of carbon dioxide has a volume of 240 mL at a temperature of 22°C. If the pressure remains constant, what is the vol

astraxan [27]

Answer:

Volume of the $CO_{2}$ gas at 44°C is <u>258 ml.</u>

Explanation

here,

using Charles' law ,

$\frac{V}{T} =\frac{v}{t}$

where , V= initial volume v= final volume

T=initial temperature t = final temperature

Given - pressure is constant ,

so , putting the values -

V= 240ml

T= 22 + 273K = 295K (since converting celsius into kelvin that

is +273K )

v =?

t = 44+ 273K = 317K

Now , putting the given values in charles' law ,

$\frac{240ml}{295K} =\frac{v}{317K}$

240ml x317K = v x 295K (through cross multiplication )

v = $\frac{240ml\times317K}{295K}$

= 258ml .

thus ,<u> the volume of carbon dioxide in a container at 44°C IS 258ml .</u>

7 0

3 years ago