Q. 24

# A plate of area 1

Answer :

Given

Surface area of the plate, *A* = 10 cm^{2} = 10 × 10^{−4} m^{2}

Thickness of copper deposited, *t* = 10 μm = 10^{−5} m

Density of copper = 9000 kg/m^{3}

Electrochemical equivalent of copper –3 × 10^{–7} kg C^{–1}

specific heat capacity of water – 4200 J kg^{–1} K^{–1}

Volume, V of copper deposited,

*V* = *A×*(2*t*)

Where

A is the area of the plates

t is the thickness of the plates

Putting the values in the above formula, we get

We know, mass of copper deposited,

*m* = Volume × Density

Using the Faraday's Laws

*m* = *ZQ*

where

m = mass of the substance

Q= charge

Z= Electrochemical equivalent

Putting the values in the above formula, we get

⇒ 18 × 10^{−5} = 3 × 10^{−7} × *Q*

⇒ *Q* = 6 × 10^{2} C

Now,

The energy spent by the cell will be = Work done by the cell

⇒*W* = *V× Q*

Where

V is the potential difference

Q is the charge

Putting the values in the above formula, we get

= 12 × 6 × 10^{2}

= 72 × 10^{2} = 7.2 kJ

Let us take ∆*θ* as the rise in temperature of water.

If this amount of energy is used to heat 100 g of water, then-

Q = c×m ×∆*θ*

Where,

c= specific heat of water

Q= Heat

m = mass of water

∆θ = change in temperature

Substituting

⇒7.2 × 10^{3} = 100 × 10^{−3} × 4200 × ∆*θ*

⇒ ∆*θ* = 17 K

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