Thermal and Chemical Effects of Electric Current
Exercises, Q1 - Q12
1. An electric current 2.0 A passes through a wire of resistance 25 Ω. How much heat will be developed in 1 minute?
ANSWER: Given, i = 2.0 A, R =25 Ω,
t =1 minute =60 s.
The heat produced in this duration,
H = i²Rt
=2²*25*60 J
=6000 J
=6.0x10³ J.
2. A coil of resistance 100 Ω is connected across a battery of emf 6·0 V. Assume that the heat developed in the coil is used to raise its temperature. If the heat capacity of the coil is 4·0 J/K, how long will it take to raise the temperature of the coil by 15°C?
ANSWER: Heat capacity of the coil =4·0 J/K which means that to raise the temperature of the coil by 1°C it requires 4·0 J. Hence to raise the temperature by 15°C, the heat required =4·0*15 =60 J.
Now given that R =100 Ω, V =6·0 volts. So the heat produced per second
=V²/R
=6²/100 J
=0·36 J.
Time required to produce 6 J heat at this rate =60/0·36 s
=6000/36 s
=1000/(6*60) minutes
=2·8 minutes.
3. The specification on a heater coil is 250 V, 500 W. Calculate the resistance of the coil. What will be the resistance of a coil of 1000 W to operate at the same voltage?
ANSWER: Let the resistance of the coil =R, Given that at the potential difference of V =250 volts the power of the coil, P =500 W.
Since the power, P =V²/R,
→R =V²/P
=250²/500 Ω
=125 Ω.
If we need to produce a power of 1000 W at the same voltage, the required resistance of the coil,
R =250²/1000 Ω
=62·5 Ω.
4. A heater coil is to be constructed with a nichrome wire (⍴ =1·0x10⁻⁶ Ω-m) which can operate at 500 W when connected to a 250 V supply. (a) What would be the resistance of the coil? (b) If the cross-sectional area of the wire is 0·5 mm², what length of the wire will be needed? (c) If the radius of each turn is 4·0 mm, how many turns will be there in the coil?
ANSWER: (a) Power, P =500 W, the potential difference across the heater coil, V =250 volts,
then P =V²/R, where R = resistance of the coil.
→R =V²/P
=250²/500 Ω
=125 Ω.
(b) Since, ⍴ =RA/l, where A is the area of cross-section and l, is the length of the wire. Given A =0·5 mm² =5·0x10⁻⁷ m². Hence the needed length of the wire,
l =RA/⍴
=125*5·0x10⁻⁷/(1·0x10⁻⁶) m
=625/10 m
=62·5 m
(c) Radius of each turn of the coil =4·0 mm (given). Length of the wire in one turn =2πr =2π*4 mm =8π mm,
Since the length of the wire is 62·5 m, the number of turns in the coil,
=62·5*1000/(8π)
=2488·10 turns
≈2500 turns.
5. A bulb with a rating of 250 V, 100 W is connected to a power supply of 220 V situated 10 m away using a copper wire of an area of cross-section 5 mm². How much power will be consumed by the connecting wires? The resistivity of copper = 1.7x10⁻⁸ Ω-m.
ANSWER: The rating is 250 V, 100 W. Hence, the resistance of the bulb,
R =V²/P
=250²/100 Ω
=625 Ω
The bulb is 10 m away from the source, so the total length of the copper connecting wire =2*10 =20 m. The resistance of this wire, r = ⍴l/A.
Given ⍴ =1·7x10⁻⁸, A =5 mm² =5x10⁻⁶ m
So, r =1·7x10⁻⁸*20/(5x10⁻⁶) Ω
=6·8x10⁻² Ω
Now the source =220 V, total resistance of the bulb and the copper wire,
R' =R +r =625 +6·8x10⁻² Ω
=625·068 Ω
Current in the circuit, i =V/R'
→i =220/625·068 =0·352 A
(Note: even neglecting the copper wire resistance for calculating the current, we get the same current)
Thus the power consumed the connecting wire =i²R'
=(0·352)²*6·8x10⁻² W
=0·0084 W
=8·4 mW.
6. An electric bulb, when connected across a power supply of 220 V, consumes a power of 60 W. If the supply drops to 180 V, what will be the power consumed? If the supply is suddenly increased to 240 V, what will be the power consumed?
ANSWER: V =220 volts, P =60 W, so the resistance of the bulb's coil,
R =V²/P
=220²/60 Ω
=806.70 Ω
When V =180 volts, power consumed by the bulb =V²/R
=180²/806·70
≈40 W.
For V =240 volts, power consumed by the bulb
=V²/R
=240²/806·70 W
=71·40 W.
7. A servo voltage stabilizer restricts the voltage output to 220 V 士1%. If an electric bulb rated at 220 V, 100 W is connected to it, what will be the minimum and maximum power consumed by it?
ANSWER: The resistance of the bulb,
R =V²/P
=220²/100 Ω
=484 Ω.
The voltage range is from 220-2·2 =217·8 volts to 220+2·2=222·2 volts. Hence the minimum power consumed,
=V²/R
=(217·8)²/484 Ω
=98 W
and the maximum power consumed,
=(222·2)²/484 Ω
=102 W.
8. An electric bulb marked 220 V, 100 W will get fused if it is made to consume 150 W or more. What voltage fluctuation will the bulb withstand?
ANSWER: The resistance of the bulb,
=V²/P
=220²/100 Ω
=484 Ω
The bulb gets fused at 150 W or more. Hence the voltage at this power
V²/R =P
→V² =PR
=150*484 =72600
→V =270 volts.
Hence the bulb will withstand voltage up to 270 V.
9. An immersion heater rated 1000 W, 220 V is used to heat 0.01 m³ of water. Assuming that the power is supplied at 220 V and 60% of the power supplied is used to heat the water, how long will it take to increase the temperature of the water from 15°C to 40°C?
ANSWER: The temperature difference of the water ΔT =40°C -15°C =25°C.
Mass of water, m = 0.01*1000 kg
=10 kg
Specific heat of water, s =4200 J/kg-K
Hence the heat required by the given quantity of water to raise the temperature from 15°C to 40°C,
=ms.ΔT
=10*4200*25 J
=1·05x10⁶ J
Given that 60% of the power supplied by the heater is used to heat the water, hence for 1·05x10⁶ J used for heating, energy supplied by the heater
=1·05x10⁶/0·60 J
=1·75x10⁶ J
Suppose the time taken by the 1000 W heater to produce this amount of energy = t.
Then, 1000*t =1·75x10⁶
→t =1·75x10³ s
≈29 minutes.
10. An electric kettle used to prepare tea, takes 2 minutes to boil 4 cups of water (1 cup contains 200 cc of water) if the room temperature is 25°C. (a) If the cost of power consumption is Rs. 1.00 per unit (1 unit = 1000 Watt-hour), calculate the cost of boiling 4 cups of water. (b) What will be the corresponding cost if the room temperature drops to 5°C?
ANSWER: (a) Amount of water =4*200 cc
=800 cc
Thus mass of water m =0.80 kg
Temperature difference,
ΔT =100°C -25°C =75°C.
Heat supplied by the kettle =ms.ΔT
=0.80*4200*75 J
=2.52x10⁵ J
1 unit =1000 W-hour
=1000*3600 J
=36x10⁵ J
Cost of 1 unit energy = Rs 1.00
Hence cost of boiling 4 cups of water,
=(2.52x10⁵)/(36x10⁵) rupees
=0.07 rupees
=7 paise.
(b) If the room temperature is 5°C, the temperature difference ΔT =100°C-5°C
=95°C
The heat needed to boil =msΔT
=0.80*4200*95 J
=3.192x10⁵ J
Now the cost of boiling,
=(3.192x10⁵)/(36x10⁵) rupees
=0.09 rupees
=9 paise.
11. The coil of an electric bulb takes 40 watts to start glowing. If more than 40 W is supplied, 60% of the extra power is converted into light and the remaining into heat. The bulb consumes 100 Watt at 220 V. Find the percentage drop in the light intensity at a point if the supply voltage changes from 220 V to 200 V.
ANSWER: Rating of the bulb -100 W at 220 V. Hence the resistance of the bulb coil R =V²/P
=220²/100 Ω
=484 Ω.
When the voltage drops to 200 V, the power consumed by the bulb,
=V²/R
=200²/484 W
=82.65 W
Above 40 W, 60% of power is converted to light. Initially, the power converted to light =0.60(100 W -40 W)
=36 W
When the voltage drops to 200 V, the power converted to light,
=0.60(82.65 W -40 W)
=25.59 W
The drop in power for light =36 -25.59 W
=10.41 W
Since the intensity of light is proportional to the power, the percentage drop in light intensity,
=10.41*100/36
=29%
12. The 2.0 Ω resistor shown in figure (33-E1) is dipped into a calorimeter containing water. The heat capacity of the calorimeter together with water is 2000 J/K. (a) If the circuit is active for 15 minutes, what would be the rise in the temperature of the water? (b) Suppose the 6.0 Ω resistor gets burnt. What would be the rise in the temperature of the water in the next 15 minutes? 
The figure for Q-12
ANSWER: (a) Heat capacity of calorimeter with water C =2000 J/K.
6 Ω and 2 Ω resistors are in parallel, its equivalent resistance,
R' =6*2/(6+2) Ω
=1·5 Ω
Other resistance in the circuit R" =1 Ω
The total resistance of the circuit,
R =R' +R"
=1·5 +1 =2·5 Ω
Current in the circuit, i =6/2·5 =2·4 A
Let the current in the 2 Ω resistor =i', then the current in the 6 Ω resistor will be =i -i'.
The voltage difference across both parallel resistors will be the same. Hence,
i'*2 =(i -i')*6
→2i' +6i' =6i
→8i' =6i
→i' =3i/4 =3*2·4/4 A
→i' =1·8 A.
If the circuit is active for 15 minutes, the heat transferred to the calorimeter,
H =i'²*2*(15*60) J
=(1·8)²*1800 J
=5832 J
The rise in the temperature of the water,
=H/C
=5832/2000 °C
=2·9°C.
(b) When the 6 Ω resistor gets burnt, the total resistance in the circuit,
R =2 Ω +1 Ω =3 Ω
Current in the circuit, i =6/3 =2 A.
Now the heat transfered to the calorimeter by the 2 Ω resistor in next 15 minutes,
H =i²Rt
=2²*2*(15*60) J
=8*900 J =7200 J
Rise in temperature of water due to this amount of heat,
=7200/2000 =3·6°C.
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CHAPTER- 33- Thermal and Chemical Effects of Electric Current
CHAPTER- 32- Electric Current in ConductorsCHAPTER- 31- CapacitorsCHAPTER- 30- Gauss's Law
CHAPTER- 29- Electric Field and Potential
CHAPTER- 28- Heat Transfer
OBJECTIVE -I
CHAPTER- 26-Laws of Thermodynamics
CHAPTER- 25-CALORIMETRY
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OBJECTIVE-I
OBJECTIVE-II
EXERCISES - Q-11 to Q-18
CHAPTER- 24-Kinetic Theory of Gases
CHAPTER- 23 - Heat and Temperature
CHAPTER- 21 - Speed of Light
CHAPTER- 20 - Dispersion and Spectra
CHAPTER- 19 - Optical Instruments
CHAPTER- 18 - Geometrical Optics
CHAPTER- 17 - Light Waves
CHAPTER- 16 - Sound Waves
CHAPTER- 15 - Wave Motion and Waves on a String
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- 33- Thermal and Chemical Effects of Electric Current
CHAPTER- 32- Electric Current in Conductors
CHAPTER- 31- Capacitors
CHAPTER- 30- Gauss's Law
CHAPTER- 29- Electric Field and Potential
CHAPTER- 28- Heat Transfer
CHAPTER- 26-Laws of Thermodynamics
CHAPTER- 25-CALORIMETRY
Questions for Short Answer
OBJECTIVE-I
OBJECTIVE-II
CHAPTER- 24-Kinetic Theory of Gases
CHAPTER- 23 - Heat and Temperature
CHAPTER- 21 - Speed of Light
CHAPTER- 20 - Dispersion and Spectra
CHAPTER- 19 - Optical Instruments
CHAPTER- 18 - Geometrical Optics
CHAPTER- 17 - Light Waves
CHAPTER- 16 - Sound Waves
CHAPTER- 15 - Wave Motion and Waves on a String
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
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CHAPTER- 5 - Newton's Laws of Motion
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CHAPTER- 4 - The Forces
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CHAPTER- 3 - Kinematics - Rest and Motion
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CHAPTER- 2 - "Physics and Mathematics"
CHAPTER- 2 - "Physics and Mathematics"
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