A cube of gold (initially at a temperature of 20.5°C) that is 6.50 mm on a side

Question

A cube of gold (initially at a temperature of 20.5°C) that is 6.50 mm on a side is connected across the terminals of a 15.0-µF capacitor that initially has a potential difference of 120.0 V between its plates.

a) What time is required to fully discharge the capacitor? The capacitor is fully discharged when the charge is less than 0.01 % which roughly corresponds to a time of 10t.

b) When the capacitor is fully discharged, what is the temperature (in °C) of the gold cube?

Speci?c heat of gold = 0.129 J/(g* C deg)
Density of gold = 19.3 g/cm3

Explanation / Answer

The side of the cube is a = 6.50 mm = 6.50 * 10^-3 m The capacitor is C = 15.0 uF = 15.0 * 10^-6 F The potential difference is E = 120.0 V a)We know from the relation Q_o = C * E Also, Q = Q_o * (1 - exp(-t/RC)) ---------(1) where R is the resistance and has value (p * a/A) (p ---- rho)(resistivity of gold) A = a^2 Here,Q = Q_o * (0.01/100) Solving equation (1) for t and substitute the above values we get the time b)The energy stored in the capacitor when it is fully discharged W = (1/2)C * E^2 = (1/2) * (Q/E) * E^2 = (1/2) * Q * E We know that W = m * s * T or T = (W/m * s) where m = d * V (m -- mass of gold cube,d -- density and V -- volume) where V = a^3,s is specific heat of gold and T is temperature Specific heat of gold = 0.129 J/(g* C deg) = 0.129 * 10^3 J/kg * oC) Density of gold = 19.3 g/cm3 = 19.3 * (10^-3/(10^-2)^3) kg/m^3

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