Answer:
The amount of heat required is [tex]H_t = 1.37 *10^{6} \ J [/tex]
Explanation:
From the question we are told that
The mass of water is [tex]m_w = 20 \ ounce = 20 * 28.3495 = 5.7 *10^2 g[/tex]
The temperature of the water before drinking is [tex]T_w = 3.8 ^oC[/tex]
The temperature of the body is [tex]T_b = 36.6^oC[/tex]
Generally the amount of heat required to move the water from its former temperature to the body temperature is
[tex]H= m_w * c_w * \Delta T[/tex]
Here [tex]c_w [/tex] is the specific heat of water with value [tex]c_w = 4.18 J/g^oC [/tex]
So
[tex]H= 5.7 *10^2 * 4.18 * (36.6 - 3.8)[/tex]
=> [tex]H= 7.8 *10^{4} \ J [/tex]
Generally the no of mole of sweat present mass of water is
[tex]n = \frac{m_w}{Z_s}[/tex]
Here [tex]Z_w[/tex] is the molar mass of sweat with value
[tex]Z_w = 18.015 g/mol[/tex]
=> [tex]n = \frac{5.7 *10^2}{18.015}[/tex]
=> [tex]n = 31.6 \ moles [/tex]
Generally the heat required to vaporize the number of moles of the sweat is mathematically represented as
[tex]H_v = n * L_v[/tex]
Here [tex]L_v[/tex] is the latent heat of vaporization with value [tex]L_v = 7 *10^{3} J/mol[/tex]
=> [tex]H_v = 31.6 * 7 *10^{3} [/tex]
=> [tex]H_v = 1.29 *10^{6} \ J [/tex]
Generally the overall amount of heat energy required is
[tex]H_t = H + H_v[/tex]
=> [tex]H_t = 7.8 *10^{4} + 1.29 *10^{6}[/tex]
=> [tex]H_t = 1.37 *10^{6} \ J [/tex]
