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2 years ago

How does a fire extinguisher works ?

Chemistry

1 answer:

Usimov [2.4K]2 years ago

8 0

Answer:

Most fire extinguishers work by separating the fuel from the oxygen. The oxygen comes from the air. It is the same oxygen we breathe. Since the oxygen has to be in contact with the fuel, if you can coat the fuel with something that keeps the oxygen away, the fire will go out. Some ways of using fire extinguisher are:

 Water extinguishers work mainly by cooling (removing heat), but a dense spray of water droplets also helps to cut off oxygen.

 Dry powder extinguishers soak up heat, melt on the fuel, and cut off oxygen, but they also neutralize the fuel (cutting it off from the fire) .

Explanation:

I hope it was helpful...

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How much energy (heat) is required to convert 248 g of water from 0oC to 154oC? Assume that the water begins as a liquid, that t

Arturiano [62]

Answer: The amount of heat required is 775.7 kJ

Explanation:

The processes involved in the given problem are:

$1.)H_2O(s)(0^oC,273K)\rightarrow H_2O(l)(0^oC,273K)\\2.)H_2O(l)(0^oC,273K)\rightarrow H_2O(l)(100^oC,373K)\\3.)H_2O(l)(100^oC,373K)\rightarrow H_2O(g)(100^oC,373K)\\4.)H_2O(g)(100^oC,373K)\rightarrow H_2O(g)(154^oC,427K)$

Now, we calculate the amount of heat released or absorbed in all the processes.

For process 1:

$q_1=m\times L_f$

where,

$q_1$ = amount of heat absorbed = ?

m = mass of water or ice = 248 g

$L_f$ = latent heat of fusion = 334 J/g

Putting all the values in above equation, we get:

$q_1=248g\times 334J/g=84832J$

For process 2:

$q_2=m\times C_{p,l}\times (T_{2}-T_{1})$

where,

$q_2$ = amount of heat absorbed = ?

$C_{p,l}$ = specific heat of water = 4.184 J/g°C

m = mass of water = 248 g

$T_2$ = final temperature = $100^oC$

$T_1$ = initial temperature = $0^oC$

Putting all the values in above equation, we get:

$q_2=248g\times 4.184J/g^oC\times (100-0)^oC=103763.2J$

For process 3:

$q_3=m\times L_v$

where,

$q_3$ = amount of heat absorbed = ?

m = mass of water or ice = 248 g

$L_v$ = latent heat of vaporization = $40.79kJ/mol\times \frac{1000}{18}=2266.1J/g$ (Conversion factor used: 1 kJ = 1000 J and molar mass of water = 18 g/mol)