Assume that a daily diet of 2000 calories (i. e. 8.37 x 106 J) is converted completely to body heat.

a) How many g of water (as sweat) would need to evaporate to cool that person off?
b) If instead of evaporating water, the heat was used to raise the temperature of some water from 25.0 °C to 35.0 °C, how much water could be heated?

(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.

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

mass in gram ⇒ 199.2857 kg x 1000 g = 199285.7 g

the answer is 1 i believe