Question

Ten moles of a gas (molar heat capacity for constant volume process is $C_V$) is enclosed in rigid hollow sphere of inner radius $a$ and outer radius $5a$ and its temperature is $3T_0$ at $t=0$. Heat is conducted out to the environment (temperature $T_0$) through the sphere material material of conductivity $K$ and negligible heat capacity. At $t=t_1$ (second) the temperature of the gas is found to be $2T_0$ then find the value of $t_1$. [All quantities are expressed in SI units and take $C_V \times (m2) = 2\pi Ka$]

Answer 1

2 ln 2

Answer 2

The thermal resistance of the spherical shell is $R_{th} = \frac{1}{4\pi K} (\frac{1}{a} - \frac{1}{5a}) = \frac{1}{5\pi K a}$. The rate of heat loss is $\frac{dQ}{dt} = - \frac{T(t) - T_0}{R_{th}} = -5\pi K a (T(t) - T_0)$. Also, $\frac{dQ}{dt} = n C_V \frac{dT}{dt}$. Thus, $n C_V \frac{dT}{dt} = -5\pi K a (T(t) - T_0)$. Let $y = T(t) - T_0$. Then $\frac{dy}{dt} = \frac{dT}{dt}$. $n C_V \frac{dy}{dt} = -5\pi K a y$. $\int_{3T_0-T_0}^{2T_0-T_0} \frac{dy}{y} = - \int_{0}^{t_1} \frac{5\pi K a}{n C_V} dt$. $\ln(2T_0) - \ln(T_0) = - \frac{5\pi K a}{n C_V} t_1$. $\ln(2) = \frac{5\pi K a}{n C_V} t_1$. $t_1 = \frac{n C_V}{5\pi K a} \ln(2)$. Given $n=10$ and $C_V \times 2 = 2\pi Ka$, so $C_V = \pi Ka$. $t_1 = \frac{10 (\pi Ka)}{5\pi K a} \ln(2) = 2 \ln(2)$.