# H C VERMA Solutions for Class 12-science Physics Chapter 11 - Thermal and Chemical Effect of Electric Current

## Chapter 11 - Thermal and Chemical Effect of Electric Current Exercise 218

An electric current of 2.0A passes through a wire of resistance 25 . How much heat will be developed in minute?

Heat=

A coil of resistance 100 is connected across a battery of emf 6.0V. 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?

Heat produced by coil=heat to increase temperature of coil

min

The specification on a heater coil is 250V, 500W. Calculate the resistance of the coil. What will be the resistance of the coil of 1000W to operate at the same voltage?

Now, P=1000W

## Chapter 11 - Thermal and Chemical Effect of Electric Current Exercise 219

A heater-coil is to be constructed with a nichrome wire () which can operate at 500W when connected to a 250V supply. (a) What would be the resistance of the coil? (b) If the cross-sectional area of the wire is, what length of the wire will be needed? (c) If the radius of each turn is 4.0mm, how many turns will be there in the coil?

(a)

(b)

m

(c) Number of turns=

A bulb with rating 250V, 100W is connected to a power supply of 220V situated 10m away using a copper wire of area pf cross-section. How much power will be consumed by the connecting wires? Resistivity of copper=.

Resistance of the bulb

Resistance of copper wire

Total resistance of copper wire(both side connecting wire) =

The effective resistance

The current supplied by the power station

Power supplied/consumed by both side connecting wire

mW

An electric bulb, when connected across a power supply of 220V, consumes a power of 60W. If the supply drops to 180V, what will be the power consumed? If the power supply is suddenly increased to 240V, what will be the power consumed?

Power=

(a)

W

(b)

W

A servo voltage stabilizer restricts the voltage output to 220V. If an electric bulb rated at 220V, 100W is connected to it, what will be the minimum and maximum power consumed by it?

Resistance of the bulb

For minimum power, voltage should be minimum

Power, W

For maximum power, voltage should be maximum

W

An electric bulb marked 220V, 100W will get fused if it is made to consume 150W or more. What voltage fluctuation will the bulb withstand?

Resistance of the bulb

Let maximum voltage bulb can withstand

volt

An immersion heater rated 1000W, 220V is used to heat 0.01 of water. Assuming that the power is supplied at 220V 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 to 40?

Power supplied by the heater=1000W

Power consumed by water=W

minutes

An electric kettle used to prepare tea, takes 2 minutes to boil 4 cups of water (1 cup contains 200 cc of water) is the room temperature is 25. (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?

(a) volume of water boiled=cc

Mass of water boiled=

Kg

Heat required for boiling water

J

We know

1000 watt-hour=

cost of boiling 4 cups of water=

(b) If room temperature drops to

Heat required,

J

cost of boiling 4 cups of water=

The coil of an electric bulb takes 40 watts to start glowing. If more than 40W is supplied, 60% of the extra power is converted into light and the remaining into heat. The bulb consumes 100W at 220V. Find the percentage drop in the light intensity at a point if the supply voltage changes from 220V to 200V.

Resistance of the bulb,

Case 1: When 220V supply is used.

Excess power=100-40=60W

Power converted into light=W

Case 2: When 200V supply is used.

Power consumed by bulb=W

Excess power=82.64-40=42.64W

Power converted into light=W

Now, percentage drop in intensity of light=

The 2.0 resistor shown in figure is dipped into a calorimeter containing water. The heat capacity of the calorimeter together with water is 2000J/K. (a) If the circuit is active for 15 minutes, what would be the rise in the temperature of the water? (b) Suppose the resistor gets burnt. What would be the rise in the temperature of the water in the next 15 minutes?

Equivalent resistance of the circuit

Current through battery

A

6 and 2 resistance are in parallel combination so current will be distributed in inverse ratio of resistance i.e., in ratio of 1:3.

Current in 2 resistance=A

(a) Heat generated by 2 resistance= heat taken by water and calorimeter

(b) Equivalent resistance of the circuit

Current in 2 resistance

I=2A

Heat generated by 2 resistance=heat taken by water and calorimeter

The temperatures of the junctions of a bismuth-silver thermocouple are maintained at 0 and 0.001. Find the thermo-emf (Seebeck emf) developed. For Bismuth-silver, V/deg and.

Emf;

V

Find the thermo-emf developed in a copper-silver thermocouple when the junctions are kept at 0 and 40. Use the data in table (33.1).

Difference in temperature ;

Emf

V

Find the neutral temperature and inversion temperature of copper-iron thermocouple if the reference junction is kept at 0. Use the data in table (33.1).

Here,

Thus, the neutral temperature

So, inversion temperature

Find the charge required to flow through an electrolyte to liberate one atom of (a) a monovalent material and (b) a divalent material.

One equivalent mass of the substance requires 96500 coulombs.

(a) For a monovalent material

Equivalent mass=molecular mass

So, atoms require 96500C

1 atom requires=

(b) For a divalent material

Equivalent mass=

So, atoms require 96500C

1 atom requires=

Find the amount of silver liberated at cathode if 0.500A of current is passed through Ag electrolyte for 1 hour. Atomic weight of silver is g/mole.

Since Ag is monoatomic.

Equivalent mass of silver=molecular mass

gm

The ECE of silver

Using,

gm

An electroplating unit plates 3.0g of silver on a brass plate in 3.0 minutes. Find the current used by the unit. The electrochemical equivalent of silver is.

A

Find the time required to liberate 1.0 liter of hydrogen at STP in an electrolytic cell by a current of 5.0A.

Mass of 1 liter hydrogen, gm

Now,

Minutes.

Two voltameters, one having a solution of silver salt and the other of a trivalent-metal salt, are connected in series and a current of 2A is maintained for 1.50 hours. It is found that 1.00g of the trivalent-metal is deposited. (a) What is the atomic weight of the trivalent metal? (b) How much silver is deposited during this period? Atomic weight of silver is 107.9 g/mole.

(a) For the trivalent metal salt

Now,

Atomic weight=

(b)

gm

A brass plate having surface area on one side is electroplated with 0.10 mm thick silver layers on both sides using a 15A current. Find the time taken to do the job. The specific gravity of silver is 10.5 and its atomic weight is 107.9 g/mol.

A

Surface area,

Thickness, t=0.1mm=cm

Volume of Ag deposited= for one side

For both sides, Mass of Ag=gm

Using,

minutes

Figure shows an electrolyte of AgCl through which a current is passed. It is observed that 2.68g of silver is deposited in 10 minutes on the cathode. Find the heat developed in the 20 resistor during this period. Atomic weight of silver is 107.9 g/mole.

gm

sec

Now,

A

Heat developed in the 20 resistor

The potential difference across the terminals of a battery of emf 12V and internal resistance drops to 10V when it is connected to a silver voltameter. Find the silver deposited at the cathode in half an hour. Atomic weight of silver is 107.9 g/mole.

Terminal potential energy, TPD=10V

Emf=12V

We know that, while discharging

A

Using,

gm

A plate of area 10 is to be electroplated with copper (density 9000 kg/) to a thickness of 10 micrometers on both sides, using a cell of 12V. Calculate the energy spent by the cell in the process of deposition. If this energy is used to heat 100g of water, calculate the rise in the temperature of the water. ECE of copper= and specific heat capacity of water=4200 J/kg-K.

Surface area

Thickness

Volume of Ag

Mass of Ag=densityvolume

kg

Using,

C

Energy spent by cell= work done by the cell

kJ

When this energy is used to heat 100gm of water, rise in temperature.

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