Tuesday, September 1, 2020

H C Verma solutions, HEAT TRANSFER, EXERCISES, Q51-Q55, Chapter-28, Concepts of Physics, Part-II

Heat Transfer


EXERCISES, Q51 - Q55


   51. A body cools down from 50°C to 45°C in 5 minutes and to 40°C in another 8 minutes. Find the temperature of the surrounding.


Answer:  Let the temperature of surrounding =T°C.

The average temp in the first case =(50+45)/2 =47.5°C 

Temperature difference =47.5-T
Rate of cooling 
=(50-45)/5 =1°C/min
From Newton's law of cooling,
dT/dt =-bA(T'-T)
→1 =-bA(47.5-T)
→bA =1/(T-47.5)

In the second case,
Rate of cooling 
=(45-40)/8 =5/8 °C/min.
Average temperature =(45+40)/2 =42.5°C.
Temperature difference =42.5-T
From Newton's law of cooling
5/8 =-bA(42.5-T)
→5/8 =(T-42.5)/(T-47.5)
{putting value of bA}
→8T-8*42.5 =5T-5*47.5
→3T =340-237.5
→T =102.5/3 ≈34°C    



 

   52. A calorimeter contains 50 g of water at 50°C. The temperature falls to 45°C in 10 minutes. When the calorimeter contains 100 g of water at 50°C, it takes 18 minutes for the temperature to become 45°C. Find the water equivalent of the calorimeter.


Answer:  Let the water equivalent of the calorimeter =m gram. Mass of water in calorimeter =50 g.

Heat lost to the surrounding 

=(m+50)*10⁻³*s*(50-45) J

=5s(m+50)*10⁻³ J

Rate of heat loss

=5s(m+50)*10⁻³/10 J/min

=5s(m+50)*10⁻⁴ J/min 

Where s = specific heat capacity of water.

For the second case, the mass of water =100 g 
Time of cooling between the same range =18 min.
Hence the rate of heat loss
=(m+100)*10⁻³*s*(50-45)/18
=5s(m+100)*10⁻³/18 J/min
The rate of heat loss in both the cases should be the same. Equating the two we get,
5s(m+100)*10⁻³/18 =5s(m+50)*10⁻⁴
→10(m+100)=18(m+50)
→8m =1000-900
→m =100/8 =12.5 g 

 



 

   53. A metal ball of mass 1 kg is heated by means of a 20 W heater in a room at 20°C. The temperature of the ball becomes steady at 50°C. (a) Find the rate of loss of heat to the surrounding when the ball is at 50°C. (b) Assuming Newton's law of cooling, calculate the rate of loss of heat to surrounding when the ball is at 30°C. (c) Assume that the temperature of the ball rises uniformly from 20°C to 30°C in 5 minutes. Find the total loss of heat to the surrounding during this period. (d) Calculate the specific heat capacity of the metal. 


Answer:  (a) Since the ball is at steady-state at 50°C, all the heat given to it is lost to the surrounding at the same rate. Rate of heat given to the ball by the heater =20 W.

Hence the rate of heat loss to the surrounding =20 W.


(b) At 50°C, rate of cooling dT/dt=20 J/s. From Newton's law of cooling,

dT/dt =-bA(T'-T)

→20=-bA(50-20)

→bA =-2/3

At 30°C, the rate of loss of heat to surrounding,

=-bA(30-20)

=(2/3)*10

=20/3 W.



(c) Average temperature during this period =(20+30)/2 =25°C.

Rate of heat loss during this period 

dT/dt=-bA(25-20)

  =(2/3)*5 J/s

  =10/3 J/s

Time =5 min =300 s
Total loss of heat during this period =(300)*(10/3)
=3000/3 J
=1000 J. 

(d) Heat given to the ball during this time =20*300
=6000 J
Heat lost during this time =1000 J
Net heat given to the ball 
Q=6000-1000 J
=5000 J
The rise in temperature, (T'-T) 
=30-20 
=10°C
Mass, m = 1 kg,
s =specific heat capacity of the metal.
Now, Q =ms(T'-T)
→5000 =1*s*10
→s =500 J/kg-K.  

   



 

   54. A metal block of heat capacity 80 J/°C placed in a room at 20°C is heated electrically. The heater is switched off when the temperature reaches 30°C. The temperature of the block rises at the rate of 2°C/s just after the heater is switched on and falls at the rate of 0.2°C/s just after the heater is switched off. Assume Newton's law of cooling to hold. (a) Find the power of the heater. (b) Find the power radiated by the block just after the heater is switched off. (c) Find the power radiated by the block when the temperature of the block is 25°C. (d) Assuming that the power radiated at 25°C represents the average value in the heating process, find the time for which the heater was kept on.


Answer:  (a) The rate of increase in temperature =2°C/s.

Power of the heater =Heat given by the heater in one second 

=(Heat capacity)*(temperature rise in a second) 

=80*2 W

= 160 W.


(b) The rate of temperature fall at 30°C =0.2°C/s.

Hence the power radiated by the block,

=80*0.2 W

=16 W.


(c) At 30°C, rate of cooling =0.2°C. From Newton's Law of cooling,

0.20=-bA(30-20)

→-bA =0.02

At 25°C, rate of cooling,

dT/dt =-bA(25-20)

      =0.02*5

      =0.10°C/s

Hence the power radiated at 25°C,

=80*0.10 W

=8 W. 


(d) Average power radiated =power radiated at 25°C =8 W.

Power of the heater =160 W

Net power recived by the block,

=160-8 =152 W.

Net heat received by the block =80*(30-20)

=800 J.

Hence the time for which the heater was kept on,

=Heat received/power received

=800/152 s

= 5.2 s.       




 

   55. A hot body placed in a surrounding of temperature θ₀ obeys Newton's law of cooling dθ/dt =-k(θ-θ₀). Its temperature at t=0 is θ₁. The specific heat capacity of the body is s and its mass is m. Find (a) the maximum heat that the body can lose and (b) the time starting from t = 0 in which it will lose 90% of this maximum heat.


Answer:  (a) Since the temperature of the surrounding is θₒ, the maximum fall in temperature will be upto θₒ. Hence the maximum heat that the body can lose, Q=ms(θ₁-θₒ).


(b) Let the temperature of the body when it loses 90% of the heat =T.

Then, 0.9Q =ms(θ₁-T)

→0.9ms(θ₁-θ₀)=ms(θ₁-T)

→T = θ₁-0.9(θ₁-θ₀)

From Newton's law of cooling,

dθ/dt =-k(θ-θ₀) 

→dt =-{1/k(θ-θ₀)}dθ

Integrating both sides within limits {t=0 to t, and temperature from θ₁ to T}

→t =-(1/k)*ln(θ-θ₀)

Put the limit on the right side

t =-(1/k)[ln(T-θ₀) -ln(θ₁-θ₀)]

=(1/k)[ln{(θ₁-θ₀)/(θ₁-0.9(θ₁-θ₀)-θₒ}

=(1/k)[ln(θ₁-θ₀)/(0.1θ₁-0.1θ₀)]

=(ln10)/k.

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Links to the Chapters











CHAPTER- 23 - Heat and Temperature






CHAPTER- 17 - Light Waves




CHAPTER- 14 - Fluid Mechanics



CHAPTER- 13 - Fluid Mechanics


CHAPTER- 12 - Simple Harmonic Motion








CHAPTER- 11 - Gravitation




CHAPTER- 10 - Rotational Mechanics




CHAPTER- 9 - Center of Mass, Linear Momentum, Collision


CHAPTER- 8 - Work and Energy

Click here for → Question for Short Answers

Click here for → OBJECTIVE-I

Click here for → OBJECTIVE-II

Click here for → Exercises (1-10)

Click here for → Exercises (11-20)

Click here for → Exercises (21-30)

Click here for → Exercises (31-42)

Click here for → Exercise(43-54)

CHAPTER- 7 - Circular Motion

Click here for → Questions for Short Answer 

Click here for → OBJECTIVE-I

Click here for → OBJECTIVE-II

Click here for → EXERCISES (1-10)

Click here for → EXERCISES (11-20)

Click here for → EXERCISES (21-30)

CHAPTER- 6 - Friction

Click here for → Questions for Short Answer

Click here for → OBJECTIVE-I

Click here for → Friction - OBJECTIVE-II

Click here for → EXERCISES (1-10)

Click here for → Exercises (11-20)

Click here for → EXERCISES (21-31)

For more practice on problems on friction solve these- "New Questions on Friction".

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CHAPTER- 5 - Newton's Laws of Motion


Click here for → QUESTIONS FOR SHORT ANSWER

Click here for→Newton's Laws of Motion,Exercises(Q.No. 13 to 27)

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CHAPTER- 4 - The Forces

The Forces-

"Questions for short Answers"    


Click here for "The Forces" - OBJECTIVE-I


Click here for "The Forces" - OBJECTIVE-II


Click here for "The Forces" - Exercises


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CHAPTER- 3 - Kinematics - Rest and Motion

Click here for "Questions for short Answers"


Click here for "OBJECTIVE-I"


Click here for EXERCISES (Question number 1 to 10)


Click here for EXERCISES (Question number 11 to 20)


Click here for EXERCISES (Question number 21 to 30)


Click here for EXERCISES (Question number 31 to 40)


Click here for EXERCISES (Question number 41 to 52)


CHAPTER- 2 - "Physics and Mathematics"

Click here for "Questions for Short Answers"


Click here for "OBJECTIVE-II"

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