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 $12V$. Calculate the energy spent by the cell in the process of deposition. If this energy is used to heat $100g$ of water, calculate the rise in the temperature of the water. $\ce{ECE}$ of copper $= 3 \times 10^{-7}kg C^1 $ and specific heat capacity of water $= 4200Jkg^1.$
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Surface area of the plate $, A = 10 \ cm^2 = 10 \times 10^{-4}m^2$
Thickness of copper deposited, $\text{t}=10\mu\text{m}=10^{-5}\text{m}$ Density of copper $= 9000 \ kg/m^3$ Volume of copper deposited $, V = A(2t) V = 10 \times 10^4 \times 2 \times 10 \times 10^6 = 2 \times 10^2 \times 10^{10} = 2 \times 10^8m^3$ Mass of copper deposited $, m = \text{Volume $\times$ Density} = 2 \times 10^{−8} \times 9000$
$ \Rightarrow m = 18 \times 10^5\ kg$ Using the formula $, m = ZQ,$
We get $, 18 \times 10^5\ kg = 3 \times 10^{7 }\times Q$
$\Rightarrow Q = 6 \times 10^2C$
Energy spent by the cell $=$ Work done by the cell
$\Rightarrow W = VQ$
$= 2 \times 6 \times 10^2$
$= 72 \times 10^2 = 7.2 kJ$
Let $\Delta\theta$ be the rise in temperature of water.
When this energy is used to heat $100g$ of water,
We have,
$72 \times 10^3$
$= 100 \times 10^{-3 }\times4200\times\Delta\theta$
$\Rightarrow\Delta\theta=17\text{K}$
Thickness of copper deposited, $\text{t}=10\mu\text{m}=10^{-5}\text{m}$ Density of copper $= 9000 \ kg/m^3$ Volume of copper deposited $, V = A(2t) V = 10 \times 10^4 \times 2 \times 10 \times 10^6 = 2 \times 10^2 \times 10^{10} = 2 \times 10^8m^3$ Mass of copper deposited $, m = \text{Volume $\times$ Density} = 2 \times 10^{−8} \times 9000$
$ \Rightarrow m = 18 \times 10^5\ kg$ Using the formula $, m = ZQ,$
We get $, 18 \times 10^5\ kg = 3 \times 10^{7 }\times Q$
$\Rightarrow Q = 6 \times 10^2C$
Energy spent by the cell $=$ Work done by the cell
$\Rightarrow W = VQ$
$= 2 \times 6 \times 10^2$
$= 72 \times 10^2 = 7.2 kJ$
Let $\Delta\theta$ be the rise in temperature of water.
When this energy is used to heat $100g$ of water,
We have,
$72 \times 10^3$
$= 100 \times 10^{-3 }\times4200\times\Delta\theta$
$\Rightarrow\Delta\theta=17\text{K}$
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