Respuesta :
Answer:
The correct answer will be "100.7 mL". The further explanation is given below.
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:
⇒ [tex]Q_{lost}=mc \Delta T[/tex]
On putting the estimated values, we get
[tex]=70\times 3480 \times 1[/tex]
[tex]=2.436\times 10^5 \ J[/tex]
Let M will be the amount of evaporated water at the temperature of 37°C.
Required heat will be:
⇒ [tex]Q_{gain}=ML_{v}[/tex]
[tex]=M(2.42\times 10^6)[/tex]
Now, Lost heat will be equal to the required amount of heat.
⇒ [tex]Q_{lost}=Q_{gain}[/tex]
[tex]2.436\times 10^5=M(2.42\times 10^6)[/tex]
On applying cross-multiplication, we get
[tex]M=\frac{2.436\times 10^5}{2.42\times 10^6}[/tex]
[tex]=0.1007 \ kg \ or \ 100.7 \ g[/tex]
Now,
⇒ [tex]V=\frac{M}{\rho}[/tex]
On putting the estimated values, we get
[tex]=\frac{1.1007}{1000}[/tex]
[tex]=100.7 \ mL[/tex]
