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when 200 grams of water cools from 50 c to 25 c the total amount of heat energy released by the water is?

Sagot :

Libby102idn
Specific Heat:
Heat Energy= Mass of substance X Specific Heat X Change in Temp. 
1. change in temp |50-25| = 25
2. specific heat of Water(H2O) = cal/g (Celsius) 1.000
heat energy= 200g X 1.000 X 25
Heat energy = 5000cal
 

Answer : The heat released by the eater is, [tex]2.1\times 10^4J[/tex][/tex]

Explanation :

Formula used :

[tex]Q=m\times c\times \Delta T[/tex]

or,

[tex]Q=m\times c\times (T_2-T_1)[/tex]

where,

Q = heat released = ?

m = mass of water = 200 g

c = specific heat of water = [tex]4.184J/g^oC[/tex]

[tex]T_1[/tex] = initial temperature  = [tex]50^oC[/tex]

[tex]T_2[/tex] = final temperature  = [tex]25^oC[/tex]

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

[tex]Q=200g\times 4.184J/g^oC\times (25-50)^oC[/tex]

[tex]Q=20920J=2.1\times 10^4J[/tex][/tex]

Therefore, the heat released by the eater is, [tex]2.1\times 10^4J[/tex][/tex]

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