11.16 A child running a temperature of 101°F is given an antipyrin (i.e. a medicine that lowers fever) which causes an increase in the rate of evaporation of sweat from his body. If the fever is brought down to 98°F in 20 minutes, what is the average rate of extra evaporation caused, by the drug. Assume the evaporation mechanism to be the only way by which heat is lost. The mass of the child is 30 kg. The specific heat of human body is approximately the same as that of water, and latent heat of evaporation of water at that temperature is about 580 cal g ^–1.

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Payal Gupta | Contributor-Level 10

5 months ago

11.16 Initial body temp of the child, $T_{1}$ = 101°F

Final body temp of the child, $T_{2}$ = 98°F

Change in temperature, $?$ T = $T_{1} - T_{2} = (101 ° F -$ 98°F) = 3 °F = $\frac{5}{9}$ (3-32) $°$ C = 1.666 $?$

Time taken t achieve this temperature, t = 20 min

Specific heat of human body = Specific heat of water, c = 1000 cal/kg/ $?$

Latent heat of evaporation of water, L = 580 cal/g

Mass of the child, m = 30 kg

The heat lost by the child is given as $? θ = m c ? T$ = 30 $\times 1000 \times 1.666$ = 49980 cal

Let $m_{1}$ be the mass of water evaporated from the child’s body in 20 mins.

Loss of heat $? θ$ = $m_{1} \times L$ = 580&nbs

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11.16 Initial body temp of the child, $T_{1}$ = 101°F

Final body temp of the child, $T_{2}$ = 98°F

Change in temperature, $?$ T = $T_{1} - T_{2} = (101 ° F -$ 98°F) = 3 °F = $\frac{5}{9}$ (3-32) $°$ C = 1.666 $?$

Time taken t achieve this temperature, t = 20 min

Specific heat of human body = Specific heat of water, c = 1000 cal/kg/ $?$

Latent heat of evaporation of water, L = 580 cal/g

Mass of the child, m = 30 kg

The heat lost by the child is given as $? θ = m c ? T$ = 30 $\times 1000 \times 1.666$ = 49980 cal

Let $m_{1}$ be the mass of water evaporated from the child’s body in 20 mins.

Loss of heat $? θ$ = $m_{1} \times L$ = 580 $m_{1}$ = $49980 c a l$

$m_{1} =$ 86.2 gm

Rate of evaporation = $\frac{m_{1}}{t}$ = 4.31 g/min

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11.16 Initial body temp of the child,  <math><msub><mrow><mrow><mi>T</mi></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow></msub></math> = 101°FFinal body temp of the child,  <math><msub><mrow><mrow><mi>T</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msub></math> = 98°FChange in temperature,  <math><mo>? </mo></math> T = <math><msub><mrow><mrow><mi>T</mi></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow></msub><mo>-</mo><msub><mrow><mrow><mi>T</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msub><mo>=</mo><mo> (</mo><mi></mi><mn>101</mn><mo>°</mo><mi mathvariant="normal">F</mi><mo>-</mo><mi mathvariant="normal"></mi></math> 98°F) = 3 °F = <math><mfrac><mrow><mrow><mn>5</mn></mrow></mrow><mrow><mrow><mn>9</mn></mrow></mrow></mfrac></math>  (3-32) <math><mo>°</mo></math> C = 1.666 <math><mi>? </mi></math>Time taken t achieve this temperature, t = 20 minSpecific heat of human body = Specific heat of water, c = 1000 cal/kg/ <math><mi>? </mi></math>Latent heat of evaporation of water, L = 580 cal/gMass of the child, m = 30 kgThe heat lost by the child is given as <math><mo>? </mo><mi>θ</mi><mo>=</mo><mi>m</mi><mi>c</mi><mo>? </mo><mi mathvariant="normal">T</mi></math> = 30 <math><mo>×</mo><mi></mi><mn>1000</mn><mi></mi><mo>×</mo><mi></mi><mn>1.666</mn></math> = 49980 calLet <math><msub><mrow><mrow><mi>m</mi></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow></msub></math> be the mass of water evaporated from the child’s body in 20 mins.Loss of heat <math><mo>? </mo><mi>θ</mi></math> = <math><msub><mrow><mrow><mi>m</mi></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow></msub><mi></mi><mo>×</mo><mi>L</mi></math> = 580 <math><msub><mrow><mrow><mi>m</mi></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow></msub></math> = <math><mn>49980</mn><mi></mi><mi>c</mi><mi>a</mi><mi>l</mi></math><math><msub><mrow><mrow><mi>m</mi></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow></msub><mo>=</mo><mi></mi></math> 86.2 gmRate of evaporation = <math><mfrac><mrow><mrow><msub><mrow><mrow><mi>m</mi></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow></msub></mrow></mrow><mrow><mrow><mi>t</mi></mrow></mrow></mfrac></math> = 4.31 g/min

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