# A child running a

Given,

Mass of the child, m = 30kg

Initial temperature of the body of the child, T1 = 101°F

Final temperature of the body of the child, T2 = 98°F

Change in temperature, ΔT = 101 – 98 = 3°F

ΔT = [3 × (5/9)] °C

ΔT = 1.667 °C

Time taken to reduce the temperature, t = 20 min

Mass of the child, m = 30 kg = 30 × 103 g

Specific heat of the human body = Specific heat of water (c) = 1000 cal kg-1 °C-1

Latent heat of evaporation of water, L = 580 cal g–1

The heat lost by the child is given as

∆θ = mc∆T

∆θ = 30 kg × 1000 cal kg-1 °C-1 × 1.667 °C = 50000 cal

Let m1 be the mass of the water evaporated from the child’s body in 20 min.

Loss of heat through water is given by

∆θ = m1L

m1 = ∆θ/L

m1 = (50000 cal)/(580 cal g–1) = 86.2 g

Hence, average rate of extra evaporation caused by the drug, Rave = m1/t

Rave = (86.2 g)/(200 min)

Rave = 4.3 g/min.

NOTE: The specific heat is the amount of heat per unit mass required to raise the temperature by one degree Celsius.

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation view all courses RELATED QUESTIONS :

Calculate the temPhysics - Exemplar

According tPhysics - Exemplar

100 g of water isPhysics - Exemplar

One day in the moPhysics - Exemplar

During summers inPhysics - Exemplar

Explain why:
<
NCERT - Physics Part-II

In an experiment NCERT - Physics Part-II

Explain why:
<
NCERT - Physics Part-II

Explain why:
<
NCERT - Physics Part-II

A child running aNCERT - Physics Part-II