Answer : The heat your body transfer must be, 25.1 kJ
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 = ?
m = mass of water = 500.0 g
c = specific heat of water = [tex]4.18J/g^oC[/tex]
[tex]T_1[/tex] = initial temperature = [tex]25.0^oC[/tex]
[tex]T_2[/tex] = final temperature = [tex]37.0^oC[/tex]
Now put all the given value in the above formula, we get:
[tex]Q=500.0g\times 4.18J/g^oC\times (37.0-25.0)K[/tex]
[tex]Q=25080J=25.1kJ[/tex]
Therefore, the heat your body transfer must be, 25.1 kJ