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Sagot :
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
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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