Question

A plate of area 10 cm² is to be electroplated with copper (density 9000 kg m⁻³) to a thickness of 10 micrometres 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⁻¹and specific heat capacity of water = 4200 J kg⁻¹.

Answer 1

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

Volume of copper deposited, V = A(2t)

V = 10 × 10⁻⁴ × 2 × 10 × 10⁻⁶
   = 2 × 10² × 10⁻¹⁰
   = 2 × 10⁻⁸ m³

Mass of copper deposited, m = Volume × Density = 2 × 10⁻⁸ × 9000

⇒ m = 18 × 10⁻⁵ kg

Using the formula, m = ZQ, we get:-

18 × 10⁻⁵ = 3 × 10⁻⁷ × Q

⇒ Q = 6 × 10² C

Energy spent by the cell = Work done by the cell

⇒W = VQ
       = 12 × 6 × 10²
       = 72 × 10² = 7.2 kJ

Let ∆θ be the rise in temperature of water. When this energy is used to heat 100 g of water, we have:-

7.2 × 10³ = 100 × 10⁻³ × 4200 × ∆θ

⇒ ∆θ = 17 K