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⁻¹ and 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