Answer:
(a) the mass of the water is 3704 g
(b) the mass of the water is 199, 285.7 g
Explanation:
Given;
Quantity of heat, H= 8.37 x 10⁶ J
Part (a) mass of water (as sweat) need to evaporate to cool that person off
Latent heat of vaporization of water, Lvap. = 2.26 x 10⁶ J/kg
H = m x Lvap.
[tex]m = \frac{H}{L._{vap}} =\frac{8.37 * 10^6.J}{2.26*10^6\ \frac{J}{kg}} = 3.704 \ kg[/tex]
mass in gram ⇒ 3.704 kg x 1000g = 3704 g
Part (b) quantity of water raised from 25.0 °C to 35.0 °C by 8.37 x 10⁶ J
specific heat capacity of water, C, 4200 J/kg.°C
H = mcΔθ
where;
Δθ is the change in temperature = 35 - 25 = 10°C
[tex]m =\frac{H}{c* \delta \theta} = \frac{8.37 *10^6}{4200*10} = 199.2857 kg[/tex]
mass in gram ⇒ 199.2857 kg x 1000 g = 199285.7 g
