Question

g what mass of water must evaporate from the skin of a 70.0 kg man to cool his body 1.00 C? The heat of vaporization of water at body temp

Answer 1

The correct answer will be "100.7 mL". The further explanation is given below.

Step-by-step explanation:

The given values are:

Temperature,

ΔT = 1°C

Mass,

m = 70 kg

c = 3.480 J/Kg.K

Amount of released heat will be:

⇒
`Q_(lost)=mc \Delta T`

On putting the estimated values, we get

`=70* 3480 * 1`

`=2.436* 10^5 \ J`

Let M will be the amount of evaporated water at the temperature of 37°C.

Required heat will be:

⇒
`Q_(gain)=ML_(v)`

`=M(2.42* 10^6)`

Now, Lost heat will be equal to the required amount of heat.

⇒
`Q_(lost)=Q_(gain)`

`2.436* 10^5=M(2.42* 10^6)`

On applying cross-multiplication, we get

`M=(2.436* 10^5)/(2.42* 10^6)`

`=0.1007 \ kg \ or \ 100.7 \ g`

Now,

⇒
`V=(M)/(\rho)`

On putting the estimated values, we get

`=(1.1007)/(1000)`

`=100.7 \ mL`