Question

A child that weighs 26.25 kg has fever of 103. 9 "F. To cool the child's fever, you want to make 85.0 °F bath. If the specific heat capacity of is 3.475 J/g°C), what mass of water is required to cool the child's body temperature a normal 98.6 °F ​

Answer 1

Initial temperature of the body of the child, T

1

=101

o

F

Final temperature of the body of the child, T

2

=98

o

F

Change in temperature, △T=[(101−98)×5/9=5/3

o

C

Time taken to reduce the temperature, t = 20 min

Mass of the child, m=30kg=30×10

3

g

Specific heat of the human body = Specific heat of water = c

=1000cal/kg/

o

C

Latent heat of evaporation of water, L=580calg

−1

The heat lost by the child is given as:

△θ=mc△T

=30×1000×(101−98)×(5/9)

=50000cal

Let m

1

be the mass of the water evaporated from the child’s body in 20 min.

Loss of heat through water is given by:

△θ=m

1

L

∴m

1

=△θ/L

=(50000/580)=86.2g

∴ Average rate of extra evaporation caused by the drug =m

1

/t

=86.2/20

=4.3g/min.

i hope it may helps you.