A child running a temperature of 101,^{circ} | vedclass

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Question

(D) Initial temperature of the child's body,$T_{1} = 101^{\circ} F$.Final temperature of the child's body,$T_{2} = 98^{\circ} F$.Change in temperature

Vedclass · Accepted Answer

$4.3$

Vedclass · Answer

(D) Initial temperature of the child's body,$T_{1} = 101^{\circ} F$.Final temperature of the child's body,$T_{2} = 98^{\circ} F$.Change in temperature,$\Delta T = (101 - 98) 	imes \frac{5}{9} = 3 	imes \frac{5}{9} = \frac{5}{3} ^{\circ} C$.Time taken to reduce the temperature,$t = 20 \; min$.Mass of the child,$m = 30 \; kg = 30,000 \; g$.Specific heat of the human body,$c = 1 \; cal \; g^{-1} \; ^{\circ} C^{-1} = 1000 \; cal \; kg^{-1} \; ^{\circ} C^{-1}$.Latent heat of evaporation of water,$L = 580 \; cal \; g^{-1}$.The heat lost by the child's body is $\Delta Q = m c \Delta T$.$\Delta Q = 30,000 \; g 	imes 1 \; cal \; g^{-1} \; ^{\circ} C^{-1} 	imes \frac{5}{3} ^{\circ} C = 50,000 \; cal$.Let $m_{1}$ be the mass of water evaporated. Since $\Delta Q = m_{1} L$,we have $m_{1} = \frac{\Delta Q}{L}$.$m_{1} = \frac{50,000}{580} \approx 86.21 \; g$.The average rate of evaporation is $\frac{m_{1}}{t} = \frac{86.21}{20} \approx 4.31 \; g/min$.Thus,the average rate of extra evaporation is $4.3 \; g/min$.

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