A plate of area 10 cm2 is to be electroplated with copper (density 9000 kg m–3) to a thickness of 10 micrometers on both sides, using a cell of 12 V. Calculate the energy spent by the cell in the process of deposition. If this energy is used to heat 100 g of water, calculate the rise in the temperature of the water, ECE of copper –3 × 10–7 kg C–1 and specific heat capacity of water – 4200 J kg–1 K–1.
Given
Surface area of the plate, A = 10 cm2 = 10 × 10−4 m2
Thickness of copper deposited, t = 10 μm = 10−5 m
Density of copper = 9000 kg/m3
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×(2t)
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 × 102 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 × 102
= 72 × 102 = 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 × 103 = 100 × 10−3 × 4200 × ∆θ
⇒ ∆θ = 17 K