Respuesta :

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

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