Answer : The heat released by the eater is, \(2.1\times 10^4J\)[/tex]
Explanation :
Formula used :
\(Q=m\times c\times \Delta T\)
or,
\(Q=m\times c\times (T_2-T_1)\)
where,
Q = heat released = ?
m = mass of water = 200 g
c = specific heat of water = \(4.184J/g^oC\)
\(T_1\) = initial temperature = \(50^oC\)
\(T_2\) = final temperature = \(25^oC\)
Now put all the given value in the above formula, we get:
\(Q=200g\times 4.184J/g^oC\times (25-50)^oC\)
\(Q=20920J=2.1\times 10^4J\)[/tex]
Therefore, the heat released by the eater is, \(2.1\times 10^4J\)[/tex]
