Question

A body, placed in vacuum, starts cooling from the initial temperature of $T_0$ K. Let $\Delta t$ be the time required to reduce the temperature to $T_0/2$ K. Then, the approximate time needed to cool the body from $T_0$ K to $T_0/3$ K is

• A. 3.5 $\Delta t$
• B. 3.7 $\Delta t$
• C. 3.9 $\Delta t$
• D. 4.1 $\Delta t$

Answer 1

Answer: (B)

3.7 $\Delta t$

Answer 2

The problem describes a body cooling in vacuum, which means heat loss occurs primarily through thermal radiation, governed by the Stefan-Boltzmann Law.

1. Formulate the cooling equation:

The rate of heat loss by radiation from a body is given by:

$$\frac{dQ}{dt} = e \sigma A T^4$$

where $e$ is the emissivity, $\sigma$ is the Stefan-Boltzmann constant, $A$ is the surface area, and $T$ is the absolute temperature of the body.

The heat lost by the body causes a decrease in its internal energy, which can be expressed as:

$$dQ = mc dT$$

where $m$ is the mass and $c$ is the specific heat capacity of the body. Since the temperature is decreasing, we have:

$$-mc \frac{dT}{dt} = e \sigma A T^4$$

Rearranging the terms to separate variables:

$$\frac{dT}{T^4} = - \frac{e \sigma A}{mc} dt$$

Let $K = \frac{e \sigma A}{mc}$ be a constant for the given body.

$$\frac{dT}{T^4} = - K dt$$

2. Integrate the equation:

Integrate both sides from an initial temperature $T_i$ to a final temperature $T_f$ over a time interval $t$:

$$\int_{T_i}^{T_f} T^{-4} dT = - K \int_0^t dt$$ $$\left[ \frac{T^{-3}}{-3} \right]_{T_i}^{T_f} = - K t$$ $$- \frac{1}{3} \left( \frac{1}{T_f^3} - \frac{1}{T_i^3} \right) = - K t$$ $$\frac{1}{3} \left( \frac{1}{T_f^3} - \frac{1}{T_i^3} \right) = K t$$

Thus, the time $t$ required for cooling is:

$$t = \frac{1}{3K} \left( \frac{1}{T_f^3} - \frac{1}{T_i^3} \right)$$

3. Apply the given conditions:

Condition 1: Time taken to cool from $T_0$ to $T_0/2$ is $\Delta t$. Here, $T_i = T_0$ and $T_f = T_0/2$.

$$\Delta t = \frac{1}{3K} \left( \frac{1}{(T_0/2)^3} - \frac{1}{T_0^3} \right)$$ $$\Delta t = \frac{1}{3K} \left( \frac{8}{T_0^3} - \frac{1}{T_0^3} \right)$$ $$\Delta t = \frac{1}{3K} \left( \frac{7}{T_0^3} \right) \quad \text{--- (Equation 1)}$$

Condition 2: Time needed to cool from $T_0$ to $T_0/3$. Let this time be $t'$. Here, $T_i = T_0$ and $T_f = T_0/3$.

$$t' = \frac{1}{3K} \left( \frac{1}{(T_0/3)^3} - \frac{1}{T_0^3} \right)$$ $$t' = \frac{1}{3K} \left( \frac{27}{T_0^3} - \frac{1}{T_0^3} \right)$$ $$t' = \frac{1}{3K} \left( \frac{26}{T_0^3} \right) \quad \text{--- (Equation 2)}$$

4. Find the relationship between $t'$ and $\Delta t$:

Divide Equation 2 by Equation 1:

$$\frac{t'}{\Delta t} = \frac{\frac{1}{3K} \left( \frac{26}{T_0^3} \right)}{\frac{1}{3K} \left( \frac{7}{T_0^3} \right)}$$ $$\frac{t'}{\Delta t} = \frac{26}{7}$$

Calculate the numerical value:

$$\frac{26}{7} \approx 3.714$$

So, $t' \approx 3.714 \Delta t$.

Comparing this value with the given options, the calculated value $3.714 \Delta t$ is closest to $3.7 \Delta t$.