a capacitor of capacitance c is fully charged by 200v battery.it is then discharged through a small coil of resistance wire embedded in a thermally insulated block of specific heat capacity 0.025j/kg/K and mass 0.1kg.if the temperature of the block rises by 0.4K. what is the value of c.
Answers
Given data :
- specific heat, s = 0.025 J/kg • K
- mass, m = 0.1 kg
- change in temperature, ∆T = 0.4 K
- voltage = 200 v
Solution : We know that energy stored in capacitor is given by U = 1/2 cv² also we know that,
Now,
⟹ U = 1/2 * C (200)²
⟹ U = 1/2 * 40000 * C
⟹ U = 20000 * C Joule
When energy is used to heat up the block. Let be the rise in temperature, then heat energy is given by Q = ms∆T
Now,
⟹ 20000 * C = Q
⟹ 20000 * C = ms∆T
⟹ 20000 * C = 0.1 * 0.025 * 0.4
⟹ 20000 * C = 0.1 * 0.01
⟹ 20000 * C = 0.001
⟹ C = 0.001/20000
⟹ C = 0.00000005 F
⟹ C = 5 * 10^( - 8 ) F
Answer : Hence capacitance C is 5 * 10^( - 8 ) F.
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