A plate of area `10 cm^2` 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 spend 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 xx 10^(-7) kg C^(-1)` and specific heat capacity of water `= 4200 J kg^(-1) K^(-1)`.

Explanation

To solve the problem step by step, we will break it down into parts:

### Step 1: Calculate the Volume of Copper Deposited
We need to find the volume of copper that will be deposited on both sides of the plate.

1. **Area of the plate (A)** = 10 cm² = \(10 \times 10^{-4}\) m² (convert cm² to m²)
2. **Thickness of copper deposited on both sides (t)** = 10 micrometres = \(10 \times 10^{-6}\) m (convert micrometres to m)
3. **Total volume (V)** deposited = Area × Thickness on both sides = \(A \times 2t\)

\[ V = (10 \times 10^{-4}) \times 2 \times (10 \times 10^{-6}) = 2 \times 10^{-8} \text{ m}^3 \]

### Step 2: Calculate the Mass of Copper Deposited
Using the volume calculated above and the density of copper, we can find the mass of copper deposited.

1. **Density of copper (ρ)** = 9000 kg/m³
2. **Mass (m)** deposited = Volume × Density

\[ m = V \times \rho = (2 \times 10^{-8}) \times 9000 = 1.8 \times 10^{-5} \text{ kg} \]

### Step 3: Calculate the Charge Sent by the Battery
Using the mass deposited and the electrochemical equivalent of copper, we can find the charge.

1. **Electrochemical equivalent (Z)** = \(3 \times 10^{-7} \text{ kg/C}\)
2. **Charge (Q)** = Mass / Electrochemical equivalent

\[ Q = \frac{m}{Z} = \frac{1.8 \times 10^{-5}}{3 \times 10^{-7}} = 6 \times 10^{2} \text{ C} \]

### Step 4: Calculate the Energy Spent by the Cell
Using the charge and the EMF of the cell, we can find the energy spent.

1. **EMF (E)** = 12 V
2. **Energy (W)** = Charge × EMF

\[ W = Q \times E = 600 \times 12 = 7200 \text{ J} = 7.2 \text{ kJ} \]

### Step 5: Calculate the Rise in Temperature of Water
Using the energy calculated above, we can find the rise in temperature of the water.

1. **Mass of water (m)** = 100 g = 0.1 kg (convert grams to kg)
2. **Specific heat capacity of water (c)** = 4200 J/(kg·K)
3. **Energy used to heat the water (W)** = 7200 J
4. **Formula**: \(W = m \cdot c \cdot \Delta T\)

Rearranging for \(\Delta T\):

\[ \Delta T = \frac{W}{m \cdot c} = \frac{7200}{0.1 \times 4200} = \frac{7200}{420} = 17.14 \text{ K} \]

### Final Answers
- Energy spent by the cell: **7.2 kJ**
- Rise in temperature of the water: **17.14 K**