Question

An electric heater of resistance $150$ $\Omega$ and a copper voltameter connected in series. If a steady current is passed through the circuit for $40$ minutes and if $0.05$ g of Cu gets deposited on the cathode, calculate the heat generated in the coil in joule (e.c.e of Cu =$3.3$ x ${10}^{-7}kg{C}^{-1})$

• A. $1212$ J
• B. $3860$ J
• C. $1430$ J
• D. $2870$ J

Answer 1

The correct option is C $1430$ J

According to 1st Law of Faraday

$m=Zit0.05=33\times {10}^{-4}\times i\times 40\times 60\frac{5}{100}=33\times {10}^{-5}\times 40\times 60\times i\frac{5}{100}=33\times \frac{1}{{10}^5}\times 40\times 60\times i\frac{5}{100}=\frac{33}{{10}^3}\times 24\times i\frac{50}{33\times 24}=i$

Heat $=i^{2}Rt$

$=\frac{50\times 50}{576\times 33\times 33}\times 150\times 40\times 60=\frac{9\times {10}^8}{627264}=\frac{9\times {10}^5\times {10}^3}{627264}=1.43\times {10}^3=1430Joules$