Q. 184.4( 9 Votes )

# A child running a

Answer :

Given,

Mass of the child, m = 30kg

Initial temperature of the body of the child, T_{1} = 101°F

Final temperature of the body of the child, T_{2} = 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 × 10^{3} 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 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

⇒ m_{1} = (50000 cal)/(580 cal g^{–1}) = 86.2 g

Hence, average rate of extra evaporation caused by the drug, R_{ave} = m_{1}/t

⇒ R_{ave} = (86.2 g)/(200 min)

⇒ R_{ave} = 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.

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