All categories

2 years ago

Similarities between conduction convection and radiation

1 answer:

8 0

Explanation:

While conduction is the transfer of heat energy by direct contact, convection is the movement of heat by actual motion of matter; radiation is the transfer of energy with the help of electromagnetic waves.

The matter is present around us, in three states, solid, liquid and gas. The conversion of matter from one state to another is termed as a change in state, that takes place due to the exchange of heat between the matter and its surroundings. So, heat is the transition of energy from one system to another, due to the difference in temperature, which occurs in three different ways, that are conduction, convection and radiation.

People often misconstrue, these forms of heat transfer but, they are based on diverse physical interaction to transfer energy. To study the difference between conduction, convection and radiation.

Hope this helped!

You might be interested in

In each case, decide if the change is a "chemical change" or a "physical change". a) A cup of household bleach changes the color

Crazy boy [7]

Answer:A cup of household bleach changes the color of your favorite T-shirt from purple to pink. - chemical change

Water vapor in your exhaled breath condenses in the air on a cold day.- physical change

Plants use carbon dioxide from the air to make sugar.- chemical change

Butter melts when placed in the Sun- physical change

Explanation:

A chemical change leads to the formation of a new substance and is not easily reversible. A physical change does not lead to the formation of a new substance and is easily reversible. Physical changes include condensation, melting,etc while a chemical change is a chemical reaction.

8 0

4 years ago

What is symbol, molecule, electrovalent compound?

jok3333 [9.3K]

Answer:

the abbreviation form of full name is called symbol.

the smallest unit of cimpound is called molecule.

8 0

3 years ago

1. In the investigation of an unknown alcohol, there was a positive Jones test and a negative Lucas test. What deductions may be

pogonyaev

Answer:

1. When observing a positive test for the jones reagent and negative for the Lucas test, it indicates that it is in the presence of a primary alcohol.

Jones reagent behaves like a strong oxidant, where it transforms the primary alcohols into carboxylic acids and the secondary alcohols into ketones. Tertiary alcohols do not react.

With the Lucas test, tertiary alcohols react immediately producing turbidity, while secondary alcohols do so in five minutes. Primary alcohols do not react significantly with Lucas reagent at room temperature.

2. No reaction (See the attached drawing)

3. (see the attached drawing)

7 0

3 years ago

If 16 moles of al react with 3 moles of S8 how many moles of Al2 S3 will be formed

Gnom [1K]

Answer:

8 moles

Explanation:

Al reacts with $S_8$ to produce $Al_2S_3$ as

$Al+S_8\rightarrow Al_2S_3$

The balanced chemical equation is

$16Al+3S_8\rightarrow 8Al_2S_3$

In the reaction, 16 moles of $Al$ react with 3 moles of $S_8$ to produce 8 moles of $Al_2S_3$ .

8 0

3 years ago

Use the given data at 500 K to calculate ΔG°for the reaction

Anton [14]

Answer : The value of $\Delta G^o$ for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

$2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0$ = standard enthalpy of formation

Now put all the given values in this expression, we get:

$\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]$

$\Delta H^o=-1035.6kJ=-1035600J$

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]$