Question

A condenser of capacitance $C$ is fully charged by a $200V$ supply. It is then discharged through a small coil of resistance wire embedded in thermally insulated block of specific heat $250J/kg-K$ and of mass $100g.$ If the temperature of the block rises by $0.4K$, then the value of $C$ is

• A. $300\mu F$
• B. $200\mu F$
• C. $400\mu F$
• D. $500\mu F$

Answer 1

Answer: (D)

Given that
$V=200V$
$m=100g=0.1kg$
s $=250Jk{g}^{-1}{K}^{-1}4$
$\Delta \theta =0.4K$
$C=?$
The energy stored in the capacitor is given by
$U=\frac{1}{2}C{V}^2$
$=\frac{1}{2}C\times (200{)}^2$
$=2C\times 10^{4}J$
This energy is used to heat up the block. Let $\Delta \theta$ be the rise in temperature, then heat energy is given by
$Q=ms\Delta \theta =0.1\times 2.50\times 0.4$
$=10J$
Now,
$2C\times 10^4=10$
$\Rightarrow C=\frac{10}{2\times 10^4}=5\times 10^{-4}$
$\therefore C=500\mu F$