Q.A pan filled with hot food cools from $94\ ^\circ\text{C}$ to $86\ ^\circ\text{C}$ in $2$ minutes when the room temperature is at $20\ ^\circ\text{C}$. How long will it take to cool from $71\ ^\circ\text{C}$ to $69\ ^\circ\text{C}$?

The key idea here is that both boxes are identical in shape, size, and surroundings, and both are filled with ice. The only difference is the material of the walls, which changes how fast heat flows in from the surroundings. Melting ice requires a fixed amount of heat per unit mass (the latent heat of fusion), so the time to melt a given mass is inversely proportional to the rate of heat transfer.

Let’s work through it step by step.

1. Heat flow through a wall For a box with walls of thickness $d$ and surface area $A$, the rate of heat conduction is given by Fourier’s law:

Here $k$ is the thermal conductivity. Since the surroundings and ice temperature are the same for both boxes, and $A$ and $d$ are identical, the heat current is directly proportional to $k$.

2. Heat required to melt ice To completely melt a mass $m$ of ice at $0^\circ\text{C}$, the heat needed is

where $L_f$ is the latent heat of fusion. For box A, the entire mass $m$ melts in 21 minutes. For box B, we only want half the mass, i.e., $m/2$, to melt.

3. Relating time to conductivity Since $Q = (dQ/dt) \times t$, we have

and for half the mass in box B:

Dividing the second equation by the first gives:

4. Plug in the numbers $k_A = 180$, $k_B = 270$, $t_A = 21$ minutes. So: $\frac{1}{2} = \frac{270}{180} \cdot \frac{t_B}{21}$ …

Concept and Intuition

A Carnot engine is an ideal heat engine that operates on a reversible cycle between two temperature reservoirs: a high-temperature source (at $T_1$) from which it absorbs heat ($Q_1$), and a low-temperature sink (at $T_2$) to which it rejects heat ($Q_2$). The engine does work ($W$) in the process.

The efficiency ($\eta$) of any heat engine is defined as the ratio of the useful work done to the heat absorbed from the source:

Since, by the first law of thermodynamics, the work done is the difference between the heat absorbed and the heat rejected ($W = Q_1 - Q_2$), the efficiency can also be written as:

For a Carnot engine, and only for a Carnot engine, there's a special relationship between the heat exchanged and the absolute temperatures of the source and sink. This relationship is a direct consequence of the second law of thermodynamics and the definition of the Kelvin temperature scale:

This means the efficiency of a Carnot engine can also be expressed solely in terms of the absolute temperatures:

This formula is crucial because it tells us the maximum possible efficiency for any heat engine operating between these two temperatures. Our strategy will be to use the given information about the heat exchange to find the ratio $Q_2/Q_1$, then equate this to $T_2/T_1$ to solve for the unknown sink temperature $T_2$.

Step-by-Step Solution

1. Understand the given information and convert units: We are told that the heat released to the sink ($Q_2$) is 22% less than the heat taken from the source ($Q_1$). This can be written mathematically as:

From this, we can find the ratio of heat rejected to heat absorbed:

The temperature of the source ($T_1$) is given as $127\,^\circ\mathrm{C}$. For thermodynamic calculations, temperatures must always be in Kelvin. …

Radioactive decay is a first-order process, meaning the rate of decay (activity) at any instant is directly proportional to the number of radioactive nuclei present at that instant. This leads to an exponential decrease in the number of nuclei and, consequently, in the activity over time.

The fundamental concept here is the law of radioactive decay, which states that the activity $A(t)$ of a sample at time $t$ is related to its initial activity $A_0$ by the formula:

This formula tells us that for every equal interval of time, the activity (or the number of undecayed nuclei) reduces by the same multiplicative factor.

Let's apply this concept to solve the problem:

1. Identify the given information and the goal:

   • Initial activity: $A_0$
   • At $t_1 = 10$ minutes, activity $A_1 = \frac{1}{\sqrt{5}} A_0$.
   • After 10 more minutes (total time $t_2 = 10 + 10 = 20$ minutes), the activity $A_2 = K A_0$.
   • We need to find the value of $K$.

2. Use the first piece of information to find the decay factor for 10 minutes:

   We know that $A(t) = A_0 e^{-\lambda t}$. For $t_1 = 10$ minutes, we have:

   $A_1 = A_0 e^{-\lambda (10)}$

   Substitute the given value for $A_1$:

   $\frac{1}{\sqrt{5}} A_0 = A_0 e^{-10\lambda}$

   Dividing both sides by $A_0$ (assuming $A_0 \neq 0$):

   $e^{-10\lambda} = \frac{1}{\sqrt{5}}$

   This expression, $e^{-10\lambda}$, represents the fraction of activity remaining after 10 minutes.

3. Use the second piece of information to set up the equation for K: …

Concept & Intuition

Newton’s law of cooling states that the rate of change of temperature of a body is proportional to the difference between its temperature and the ambient (surrounding) temperature. For small temperature intervals, we can approximate the law as:

where $T_s$ is the surrounding temperature, $k$ is a constant, and we use the average temperature during the interval. This avoids calculus and works well for the problem.

Step-by-step solution

1. Set up the first interval (60 °C → 50 °C in 10 min) Average temperature: $\frac{60+50}{2} = 55$ °C. Temperature drop: $60-50 = 10$ °C. Newton’s law gives:

2. Set up the second interval (50 °C → 40 °C in 15 min) Average temperature: $\frac{50+40}{2} = 45$ °C. Temperature drop: $50-40 = 10$ °C.

3. Find the surrounding temperature $T_s$ Divide equation (1) by equation (2):

Cross-multiply:

So $135 - 110 = 3T_s - 2T_s$ → $25 = T_s$.

Thus the surroundings are at $25$ °C.

4. Find the constant $k$ …

A pendulum clock keeps time by counting the oscillations of a pendulum. The accuracy of the clock depends directly on the period of oscillation of the pendulum.

Concept and Intuition

1. Period of a Simple Pendulum: The period $T_p$ of a simple pendulum (the time for one complete swing) is given by the formula $T_p = 2\pi \sqrt{\frac{L}{g}}$, where $L$ is the length of the pendulum and $g$ is the acceleration due to gravity. From this formula, we can see that if the length $L$ increases, the period $T_p$ also increases.
2. Thermal Expansion: Materials expand when heated and contract when cooled. This phenomenon is called thermal expansion. For a linear object like a pendulum rod, the change in length $\Delta L$ due to a temperature change $\Delta T$ is given by $\Delta L = L_0 \alpha \Delta T$, where $L_0$ is the original length and $\alpha$ is the coefficient of linear expansion.
3. Connecting the Concepts: In this problem, the clock is manufactured at $32^\circ \mathrm{C}$ but operates at $47^\circ \mathrm{C}$. Since the operating temperature is higher than the manufacturing temperature, the steel pendulum rod will expand. This increase in length will cause the pendulum's period to increase.
4. Effect on Clock: If the period of the pendulum increases, each swing takes longer than it should. Consequently, the clock will "fall behind" and run slow.

Step-by-step Solution

1. Calculate the change in temperature ($\Delta T$):

   The manufacturing temperature is $T_1 = 32^\circ \mathrm{C}$.

   The operating temperature is $T_2 = 47^\circ \mathrm{C}$.

   The change in temperature is $\Delta T = T_2 - T_1 = 47^\circ \mathrm{C} - 32^\circ \mathrm{C} = 15^\circ \mathrm{C}$.

2. Determine the fractional change in the pendulum's length ($\frac{\Delta L}{L}$):

   Due to thermal expansion, the length of the pendulum changes. The fractional change in length is given by:

   $\frac{\Delta L}{L} = \alpha \Delta T$

   Given the coefficient of linear expansion for steel $\alpha = 12 \times 10^{-6} / ^\circ \mathrm{C}$.

   $\frac{\Delta L}{L} = (12 \times 10^{-6} / ^\circ \mathrm{C}) \times (15^\circ \mathrm{C})$

   $\frac{\Delta L}{L} = 180 \times 10^{-6}$

3. Relate the fractional change in length to the fractional change in the period ($\frac{\Delta T_p}{T_p}$):

   The period of a simple pendulum is $T_p = 2\pi \sqrt{\frac{L}{g}}$.

   For small changes in length, the fractional change in the period is related to the fractional change in length by:

   $\frac{\Delta T_p}{T_p} = \frac{1}{2} \frac{\Delta L}{L}$

   This relationship comes from differentiating the period formula with respect to length, or by using a binomial approximation for small changes:

   [!PROOF]

   Let $T_p = C \sqrt{L}$, where $C = 2\pi/\sqrt{g}$ is a constant.

   If $L$ changes to $L + \Delta L$, then $T_p$ changes to $T_p + \Delta T_p$.

   $T_p + \Delta T_p = C \sqrt{L + \Delta L} = C \sqrt{L \left(1 + \frac{\Delta L}{L}\right)}$

   $T_p + \Delta T_p = C \sqrt{L} \left(1 + \frac{\Delta L}{L}\right)^{1/2}$

   Since $C \sqrt{L} = T_p$, we have:

   $T_p + \Delta T_p = T_p \left(1 + \frac{\Delta L}{L}\right)^{1/2}$

   For small values of $x$, the binomial approximation states $(1+x)^n \approx 1+nx$. Here, $x = \frac{\Delta L}{L}$ and $n = \frac{1}{2}$.

   So, $T_p + \Delta T_p \approx T_p \left(1 + \frac{1}{2} \frac{\Delta L}{L}\right)$

   $T_p + \Delta T_p \approx T_p + T_p \frac{1}{2} \frac{\Delta L}{L}$

   Subtracting $T_p$ from both sides gives:

   $\Delta T_p \approx T_p \frac{1}{2} \frac{\Delta L}{L}$ …

Surface area of the spherical shell: $A = 4\pi r^2 = 4\pi (1)^2 = 12.57\text{ m}^2$.

Rate of heat flow through the steel wall (thickness $d = 5\text{ mm} = 5\times10^{-3}\text{ m}$, $\Delta T = 100\text{ K}$):

Mass of ice: volume $= \tfrac{4}{3}\pi r^3 = 4.19\text{ m}^3$, density $= 0.9\text{ g/cm}^3 = 900\text{ kg/m}^3$, so

$m = 4.19 \times 900 = 3770\text{ kg}.$ …

The concept first — separate absorption from net exchange

Stefan–Boltzmann law: a black body of surface area $A$ at absolute temperature $T$ radiates energy at the rate

Students often reach for the net exchange formula $P_{\text{net}} = \sigma A (T_1^4 - T_2^4)$. But read the question carefully: it asks for the rate of absorption of radiant energy by the outer sphere — not the net gain. Absorption and emission are two independent streams:

• What the outer sphere absorbs = whatever radiation arrives at it and is not reflected.
• What the outer sphere emits = $\sigma (4\pi(4R)^2)(2T)^4$, an entirely separate stream that leaves it.

Since the outer sphere is a black body, its absorptivity $a = 1$: it absorbs everything incident on it.

Step-by-step

1. Radiation emitted by the inner sphere. Surface area $A_{\text{in}} = 4\pi R^2$, temperature $T$:

2. Where does it go? The outer sphere is concentric and completely surrounds the inner one, so no radiation escapes the cavity. Every joule leaving the inner sphere is incident on the inner surface of the outer sphere. …

Newton’s law of cooling says that the rate of heat loss of a body is proportional to the temperature difference between the body and its surroundings. For small temperature intervals, we can approximate the cooling as exponential, leading to a simple logarithmic relation between temperature differences and time. The key insight: if we take the natural log of the temperature excess above room temperature, it decreases linearly with time. That lets us set up two equations from the two cooling intervals and solve for the unknown room temperature.

1. Set up variables and the cooling law Let the room temperature be $T_0$ (in kelvin). Newton’s law in integral form for a small interval gives

where $T_i$ and $T_f$ are initial and final temperatures, $t$ is the time, and $k$ is a positive constant depending on the liquid and container.

2. Apply to the first interval From $368\,\text{K}$ to $358\,\text{K}$ in $22$ minutes:

3. Apply to the second interval From $358\,\text{K}$ to $353\,\text{K}$ in $12.5$ minutes:

4. Eliminate $k$ by dividing the equations Divide (1) by (2):

5. Solve for $T_0$ Let $x = T_0$. Then

Exponentiate both sides:

This is messy, but we can test the given options. The room temperature is likely near $300\,\text{K}$ (about $27^\circ\text{C}$). Try $x = 300.5\,\text{K}$ (which is $27.5^\circ\text{C}$):

• First ratio: $\frac{368 - 300.5}{358 - 300.5} = \frac{67.5}{57.5} \approx 1.1739$ …

Concept and Intuition

We are asked: at what temperature do the Fahrenheit and Kelvin scales show the same numerical value? This is not a trick—it’s a straightforward conversion problem. The two scales have different zero points and different step sizes (degrees). To find where they coincide, we set the reading on one scale equal to the reading on the other, using the standard conversion formula between Fahrenheit and Kelvin.

The classic pitfall is forgetting that Kelvin has no degree symbol and that the conversion involves both a scaling factor and an offset. The formula connecting Fahrenheit ($F$) and Kelvin ($K$) is:

We want $F = K$. So we set:

Now solve for $F$.

Step-by-Step Solution

1. Set up the equation We want the same reading on both scales, so let $T$ be that common number. Then:

2. Eliminate the fraction Multiply both sides by 9 to clear the denominator:

3. Combine like terms

4. Solve for $T$ $T = \frac{2298.35}{4} = 574.5875$ …

The key insight is that mechanical energy in a damped oscillator (like a spring-mass system with damping) is proportional to the square of the amplitude. This is because energy depends on the square of displacement or velocity — for a simple harmonic oscillator, the total energy is $E = \frac{1}{2} k A^2$ at any instant, even as the amplitude decays. So if the amplitude decays as $A(t) = A_0 e^{-bt/2m}$, then energy decays as $E(t) \propto [A(t)]^2 = A_0^2 e^{-bt/m}$.

We want the time when $E(t) = \frac{1}{4} E_0$. That means the square of the amplitude has fallen to one-fourth of its initial value, which is equivalent to the amplitude itself falling to one-half (since $\sqrt{1/4} = 1/2$). Let’s work it through.

1. Write the energy decay relation directly from the amplitude law:

where $E_0$ is the initial mechanical energy. This comes from $E \propto A^2$, so the exponent doubles: $2 \times (-bt/2m) = -bt/m$.

2. Set the condition $E(t) = \frac{1}{4} E_0$:

Cancel $E_0$ (non-zero):

3. Take the natural logarithm on both sides:

Since $\ln(1/4) = -\ln 4 = -2 \ln 2$, we get:

4. Cancel the negative signs and solve for $t$: $t = \frac{2m \ln 2}{b}$ …

Concept and Intuition

When two rods of different materials are heated, they expand by different amounts because their coefficients of linear expansion differ. The problem says the steel rod is always 4 cm longer than the copper rod at any temperature. That means the difference in their lengths does not change when temperature changes. For this to happen, the increase in length of the steel rod must exactly equal the increase in length of the copper rod when both are heated by the same amount. If one expanded more than the other, the gap would widen or shrink. So we set the expansions equal and solve for the ratio of original lengths.

Step-by-step solution

1. Define variables Let $L_s$ be the length of the steel rod and $L_c$ the length of the copper rod at some initial temperature. The problem states that at any temperature,

This difference is constant.

2. Express the change in length When the temperature changes by $\Delta T$, the new lengths are:

where $\alpha_s = 1.1 \times 10^{-5} /^\circ\text{C}$ and $\alpha_c = 1.7 \times 10^{-5} /^\circ\text{C}$.

3. Apply the constant-difference condition At the new temperature, the difference must still be 4 cm:

Substitute the expressions:

4. Use the original difference We know $L_s - L_c = 4$. Expand the left side:

Since $L_s - L_c = 4$, this becomes:

5. Solve for the condition Subtract 4 from both sides: …