Question

A highly conducting uniform sphere of thermal capacity $C$ is heated by an electric heater, a resistance $R$ fitted within the sphere. A constant current $I$ is passed through the heater starting at time $t=0$ which gives constant power. The sphere loses heat at a rate equal to $k$ times the temperature difference between the sphere and the surrounding. The initial temperature of the sphere and that of the surrounding is $0^{\circ}C$. The time at which sphere attains half of its maximum attainable temperature is $\frac{C}{k} m2^a$. Then find the value of $|a|$.

Answer 1

2

Answer 2

The rate of change of temperature $T$ of the sphere is given by the heat balance equation: $C \frac{dT}{dt} = P - k(T - T_s)$ Given $T(0) = T_s = 0$, and $P = I^2R$. The equation becomes: $C \frac{dT}{dt} = P - kT$ This is a first-order linear differential equation: $\frac{dT}{dt} + \frac{k}{C}T = \frac{P}{C}$ The maximum attainable temperature $T_{max}$ is when $\frac{dT}{dt} = 0$, so $T_{max} = \frac{P}{k}$. The solution with $T(0) = 0$ is: $T(t) = T_{max} \left( 1 - e^{-\frac{k}{C}t} \right)$ We need to find $t$ when $T(t) = \frac{1}{2} T_{max}$: $\frac{1}{2} T_{max} = T_{max} \left( 1 - e^{-\frac{k}{C}t} \right)$ $\frac{1}{2} = 1 - e^{-\frac{k}{C}t}$ $e^{-\frac{k}{C}t} = \frac{1}{2}$ Taking the natural logarithm: $-\frac{k}{C}t = \ln\left(\frac{1}{2}\right) = -\ln(2)$ $t = \frac{C}{k} \ln(2)$ The problem states the time is $\frac{C}{k} m2^a$. Comparing, we have $\ln(2) = m2^a$. Assuming the question intended a form like $\frac{C}{k} \ln(a)$ or a similar structure that yields a simple integer for $a$, and given the context of such problems, it's likely that the form should lead to $a=2$. If we interpret $m2^a$ as $\ln(2)$, and assume $m=1$, then $\ln(2) = 2^a$, which gives $a = \log_2(\ln(2))$, not a simple integer. A more plausible interpretation given the options is that the form was intended to be $\frac{C}{k} \ln(a)$, leading to $a=2$. Therefore, $|a| = |2| = 2$.