Ask & Answer Question

physics ncert solutions class 11th Physics Class 11th Thermal Properties of Matter

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.

3 Views | Posted 4 months ago

Asked by Shiksha User

1 Answer

Answered by

Payal Gupta | Contributor-Level 10

4 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

...more

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

less

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

0 Upvotes 0 Downvotes

Similar Questions for you

At what temperature a gold ring of diameter 6.230cm be heated so that it can be fitted on a wooden bangle of diameter. 6.241cm? Both the diameters have been measured at room temperature (27°C).

(Given : coefficient of linear thermal expansion of gold a_L = 1.4 × 10^-⁵ K^-¹)

alok kumar singh

According to question, we can write

$d = d_{0} (1 + α Δ T) \Rightarrow 6.241 = 6.230 (1 + 1.4 \times 1 0^{- 5} \times Δ T)$

$\Rightarrow T = 126.18 + 27 = 153.18 ° C$

A copper block of mass 5.0kg is heated to a temperature of $500 ° C$ and is placed on a large ice block. What is the maximum amount of ice that can melt?

[Specific heat of copper : 0.39 J g^-1 ⁰C^-1 and latent heat of fusion of water : 335 J g^-1]

Raj Pandey

Heat Released by block = Heat gain by large Ice block

$\Rightarrow m c Δ T = M_{i c e} L$

5 × 0.39 × 500 = m_ice × 335

$m_{i c e} = \frac{5 \times 0.39 \times 500}{335}$

= 2.91 kg

A solid metallic cube having total surface area 24 m² is uniformly heated. if its temperature is increased by 10⁰C, calculate the increase in volume of the cube.

(Given a = 5.0 × 10^-4 ⁰C^-1).

Raj Pandey

Given

6a² = 24

a2 = 4

a = 2m

$Δ v = v (γ) Δ T$

$= 3 α v Δ T$

$= 3 \times 5 \times 1 0^{- 4} \times (a^{3}) \times 10$

$= 3 \times 5 \times 1 0^{- 4} \times 8 \times 10$

$\begin{array}{l} = 120 \times 1 0^{3} \\ = 1.2 \times 1 0^{5} c m^{3} \end{array}$

A glass full of hot milk is poured on the table. It begins to cool gradually. Which of the following is correct?

(a) The rate of cooling is constant till milk attains the temperature of the surrounding

(b) The temperature of milk falls off exponentially with time

(c) While cooling, there is a flow of heat from milk to the surrounding as well as from surrounding to the milk but the net flow of heat is from milk to the surrounding and that is why it cools

(d) All three phenomenon, conduction, convection and radiation are responsible for the loss of heat from milk to the surroundings

Payal Gupta

This is a multiple choice answer as classified in NCERT Exemplar

(b), (c), (d) When the hot milk in the table is transferred to the surroundings by conduction, convection and radiation.

According to newton's law of cooling temperature of the milk falls of exponentially. Heat also will be transferred from surroundings to the milk but will be lesser than that of transferred from milk to surroundings. So option b, c, d satisfy.

Refer to the plot of temperature versus time (Fig) showing the changes in the state of ice on heating (not to scale) Which of the following is correct?

(a) The region AB represents ice and water in thermal equilibrium

(b) At B water starts boiling

(c) At C all the water gets converted into steam

(d) C to D represents water and steam in equilibrium at boiling point

Payal Gupta

This is a multiple choice answer as classified in NCERT Exemplar

(a), (d) During the process AB temperature of the system is 0⁰C . hence it represents phase change that is transformation of ice into water while temperature remains 0⁰C.

BC represents rise in temperature of water from 0⁰C to 100⁰C. now water starts converting into steam which is represented by CD.

Which is only represents by a, d options

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

65k Colleges
1.2k Exams
688k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

1 Answer

Similar Questions for you

Learn more about...

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question