Q.A sphere of $0.047\ \text{kg}$ aluminium is placed for sufficient time in a vessel containing boiling water, so that the sphere is at $100\ ^\circ\text{C}$. It is then immediately transferred to $0.14\ \text{kg}$ copper calorimeter containing $0.25\ \text{kg}$ water at $20\ ^\circ\text{C}$. The temperature of water rises and attains a steady state at $23\ ^\circ\text{C}$. Calculate the specific heat capacity of aluminium.

Showing the 12 most recent of 43 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.Water of mass $3\, \text{kg}$ in a kettle of mass $1\, \text{kg}$ at an initial temperature of $30\, ^\circ \text{C}$ is heated by a heater of power $2\, \text{kW}$. When the lid of the kettle is open, heat is lost at a constant rate of $250\, \text{J}\,\text{s}^{-1}$. If the specific heat capacity of the material of the kettle is half of the specific heat capacity of water, then the time required in minutes to increase the temperature of the water to $80\, ^\circ \text{C}$ with the lid of the kettle open is (Specific heat capacity of water $= 4200\, \text{J}\,\text{kg}^{-1}\,\text{K}^{-1}$) (A) 13 (B) 7 (C) 9 (D) 21

   ›Reveal solutionSolution

   The net power available to heat the water and the kettle is the heater power minus the constant heat loss. Using the total heat capacity of the system, the required time is found to be 7 minutes.

   The key idea here is that the heater supplies energy at a fixed rate, but some of that energy is continuously lost to the surroundings. Only the remaining power actually raises the temperature of both the water and the kettle. Since the kettle itself also absorbs heat, we must account for its thermal mass using its specific heat capacity, which is given relative to water.

   Let’s work through it step by step.

   1. Find the specific heat capacity of the kettle material. The problem states it is half that of water. Water’s specific heat is $c_w = 4200 \, \text{J kg}^{-1} \text{K}^{-1}$. So the kettle’s specific heat is

   $c_k = \frac{1}{2} \times 4200 = 2100 \, \text{J kg}^{-1} \text{K}^{-1}.$

   2. Calculate the total heat capacity of the system (water + kettle). Heat capacity is mass times specific heat. For water: $m_w c_w = 3 \times 4200 = 12600 \, \text{J K}^{-1}$. For the kettle: $m_k c_k = 1 \times 2100 = 2100 \, \text{J K}^{-1}$. Total heat capacity:

   $C_{\text{total}} = 12600 + 2100 = 14700 \, \text{J K}^{-1}.$

   3. Determine the temperature rise needed. From $30^\circ \text{C}$ to $80^\circ \text{C}$, the change is

   $\Delta T = 80 - 30 = 50 \, \text{K}.$

   4. Compute the total heat energy required to raise the temperature. Using $Q = C_{\text{total}} \Delta T$:

   $Q = 14700 \times 50 = 735000 \, \text{J}.$

   5. Find the net power available for heating. …

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2. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.A tank of height 5 m is completely filled with water and a cube of side 1 cm and density $1.5\, \text{g cm}^{-3}$ is placed at the bottom of the tank. The work to be done to lift the cube at the bottom to a height of 15 m above the surface of the water is (Neglect the side of the cube when compared with the height and take acceleration due to gravity $= 10\, \text{ms}^{-2}$) (A) 25 mJ (B) 300 mJ (C) 250 mJ (D) 225 mJ

   ›Reveal solutionSolution

   The work required is the sum of the work against gravity and the work against buoyancy while the cube is underwater, plus the work against gravity alone above the water. The final result is 250 mJ, option (C).

   ---

   Concept and Intuition

   When you lift an object underwater, you don't have to fight its full weight — buoyancy helps you. The net force you must overcome while the cube is submerged is its apparent weight: actual weight minus buoyant force. Once the cube breaks the surface, you're lifting its full weight. So the total work splits into two phases:

   1. From the bottom of the tank (5 m below the surface) up to the surface (0 m relative to the water).
   2. From the surface up to 15 m above the water.

   We'll compute the work in each phase and add them.

   ---

   Step-by-step solution

   1. Convert units and find the cube's volume and mass

   Side of cube = 1 cm = $0.01\,\text{m}$.

   Volume $V = (0.01)^3 = 10^{-6}\,\text{m}^3$.

   Density $\rho_{\text{cube}} = 1.5\,\text{g/cm}^3 = 1500\,\text{kg/m}^3$.

   Mass $m = \rho_{\text{cube}} \cdot V = 1500 \times 10^{-6} = 1.5 \times 10^{-3}\,\text{kg} = 1.5\,\text{g}$.

   2. Forces on the cube underwater

   Weight: $W = mg = 1.5 \times 10^{-3} \times 10 = 0.015\,\text{N}$.

   Buoyant force: $F_b = \rho_{\text{water}} V g = 1000 \times 10^{-6} \times 10 = 0.01\,\text{N}$.

   Net downward force (apparent weight) = $W - F_b = 0.015 - 0.01 = 0.005\,\text{N}$.

   So you must apply an upward force of $0.005\,\text{N}$ to lift it at constant speed underwater.

   3. Work done underwater (from bottom to surface)

   The cube starts at the bottom of the tank, 5 m below the surface. It must be lifted to the surface — a vertical distance of 5 m.

   Work = force × distance = $0.005\,\text{N} \times 5\,\text{m} = 0.025\,\text{J} = 25\,\text{mJ}$.

   4. Work done above water (from surface to 15 m above) …

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3. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.Steam at a temperature of $100~^\circ\mathrm{C}$ is passed into water of mass $90~\mathrm{g}$ such that the temperature of the water increases from $20~^\circ\mathrm{C}$ to $40~^\circ\mathrm{C}$. Then the total mass of the water at $40~^\circ\mathrm{C}$ is (A) $3~\mathrm{g}$ (B) $93~\mathrm{g}$ (C) $30~\mathrm{g}$ (D) $120~\mathrm{g}$

   ›Reveal solutionSolution

   The key idea is that steam condenses and then cools, releasing heat that warms the water; the total mass at the end is the original water plus the condensed steam. The correct answer is 93 g.

   Concept and intuition:

   When steam at 100 °C is passed into cooler water, the steam first condenses into water at 100 °C (releasing its latent heat), and then that condensed water cools from 100 °C to the final temperature (releasing sensible heat). Both amounts of heat go into warming the original water from 20 °C to 40 °C. The total mass of water at the end is simply the original 90 g plus the mass of steam that condensed.

   Step-by-step reasoning:

   1. Identify the heat exchanges Let $m$ grams of steam condense.
      • Heat released when steam at 100 °C condenses to water at 100 °C:

   $Q_1 = m \cdot L$

    where $ L = 540 \, \text{cal/g} $ (latent heat of vaporization of water).  
   

   • Heat released when that condensed water cools from 100 °C to 40 °C:

   $Q_2 = m \cdot c \cdot (100 - 40) = m \cdot 1 \cdot 60 = 60m \, \text{cal}$

    (specific heat capacity of water $ c = 1 \, \text{cal/g}^\circ\text{C} $).
   

   2. Heat gained by the original water

   Original water mass = 90 g, temperature rise from 20 °C to 40 °C:

   $Q_{\text{gain}} = 90 \cdot 1 \cdot (40 - 20) = 90 \cdot 20 = 1800 \, \text{cal}$

   3. Apply conservation of energy Heat lost by steam = Heat gained by water:

   $Q_1 + Q_2 = 1800$

   $540m + 60m = 1800$

   $600m = 1800$ …

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4. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.If the numerical value of the temperature of a body on Fahrenheit scale is 80 more than its numerical value on Celsius scale, then the temperature of the body is (A) $120~^\circ \mathrm{F}$ (B) $120~^\circ \mathrm{C}$ (C) $60~^\circ \mathrm{F}$ (D) $60~^\circ \mathrm{C}$

   ›Reveal solutionSolution

   We use the standard conversion formula between Celsius ($C$) and Fahrenheit ($F$) scales, $F = \frac{9}{5}C + 32$, along with the given condition that the numerical value of $F$ is 80 more than $C$ (i.e., $F = C + 80$). Solving these equations simultaneously yields the temperature as $60~^\circ \mathrm{C}$.

   When dealing with temperature, it's crucial to understand that different scales (like Celsius, Fahrenheit, and Kelvin) measure the same physical quantity but assign different numerical values based on their reference points and divisions. The key to solving problems involving different scales is to use the correct conversion formula that relates them.

   For Celsius and Fahrenheit scales, there's a linear relationship. The Celsius scale sets the freezing point of water at $0~^\circ \mathrm{C}$ and the boiling point at $100~^\circ \mathrm{C}$. The Fahrenheit scale sets these points at $32~^\circ \mathrm{F}$ and $212~^\circ \mathrm{F}$ respectively. This means a $100$-degree interval on the Celsius scale corresponds to a $180$-degree interval on the Fahrenheit scale ($212 - 32 = 180$). This ratio, $\frac{180}{100} = \frac{9}{5}$, is central to the conversion.

   The relationship between temperature in Celsius ($C$) and Fahrenheit ($F$) is given by:

   $F = \frac{9}{5}C + 32$

   We are given a specific condition: the numerical value of the temperature on the Fahrenheit scale is 80 more than its numerical value on the Celsius scale. We can express this condition as an algebraic equation and then solve it simultaneously with the conversion formula.

   1. Define variables and state the conversion formula:

      Let $C$ be the numerical value of the temperature in degrees Celsius ($^\circ \mathrm{C}$).

      Let $F$ be the numerical value of the temperature in degrees Fahrenheit ($^\circ \mathrm{F}$).

      The standard conversion formula is:

      $F = \frac{9}{5}C + 32 \quad \text{(Equation 1)}$

   2. Formulate the given condition as an equation:

      The problem states that the numerical value on the Fahrenheit scale is 80 more than its numerical value on the Celsius scale. This can be written as:

      $F = C + 80 \quad \text{(Equation 2)}$

   3. Substitute and solve for the Celsius temperature:

      Now we have two equations for $F$. We can substitute Equation 2 into Equation 1 to eliminate $F$ and solve for $C$:

      $C + 80 = \frac{9}{5}C + 32$

      To solve for $C$, we gather all terms involving $C$ on one side and constant terms on the other: …

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5. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.If a solid sphere of mass 2 kg is rolling without slipping on a surface with a velocity of $10~\mathrm{ms}^{-1}$, then the total kinetic energy of the sphere is (A) 70 J (B) 140 J (C) 280 J (D) 350 J

   ›Reveal solutionSolution

   For a rolling rigid body, total kinetic energy is the sum of translational and rotational parts: $K = \frac12 mv^2 + \frac12 I \omega^2$. For a solid sphere, $I = \frac25 mR^2$ and rolling without slipping gives $\omega = v/R$, so $K = \frac12 mv^2 + \frac12 \cdot \frac25 mR^2 \cdot (v/R)^2 = \frac{7}{10} mv^2$. With $m=2$ kg and $v=10$ m/s, $K = \frac{7}{10} \cdot 2 \cdot 100 = 140$ J. The correct option is (B).

   ---

   Concept & Intuition

   When an object rolls without slipping, its motion is a combination of translation (the center of mass moving) and rotation (spinning about the center). The no-slip condition ties the two together: the point of contact is instantaneously at rest, so the angular speed $\omega$ and linear speed $v$ are related by $\omega = v/R$. The total kinetic energy is not just $\frac12 mv^2$ — we must add the rotational kinetic energy $\frac12 I \omega^2$. For a solid sphere, the moment of inertia is $I = \frac25 mR^2$, which is smaller than a hoop or disk, so the rotational contribution is a specific fraction of the translational part.

   ---

   Step-by-step solution

   1. Write the total kinetic energy formula For any rigid body rolling without slipping,

   $K = K_{\text{trans}} + K_{\text{rot}} = \frac12 m v^2 + \frac12 I \omega^2.$

   2. Apply the no-slip condition Rolling without slipping means the point of contact has zero velocity relative to the surface, giving

   $\omega = \frac{v}{R}.$

   3. Insert the moment of inertia for a solid sphere For a solid sphere of mass $m$ and radius $R$,

   $I = \frac{2}{5} m R^2.$

   4. Substitute into the rotational term

   $K_{\text{rot}} = \frac12 \left( \frac{2}{5} m R^2 \right) \left( \frac{v}{R} \right)^2 = \frac12 \cdot \frac{2}{5} m R^2 \cdot \frac{v^2}{R^2} = \frac{1}{5} m v^2.$

   5. Add translational and rotational parts …

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6. TG EAPCET 2026Set ap-2026-05-04-AN1 markMCQ
   Q.Two liquids A and B of masses ‘m’ and ‘2m’ at temperatures 30 $^\circ$C and 50 $^\circ$C respectively are mixed in a vessel of mass ‘5m’ which is at a temperature of 20 $^\circ$C. If the ratio of the specific heat capacities of the liquids A and B is 1:2 and the specific heat capacity of the material of the vessel is 0.3 times the specific heat capacity of liquid A, then the resultant temperature of the mixture is (A) 40 $^\circ$C (B) 35 $^\circ$C (C) 25 $^\circ$C (D) 38 $^\circ$C

   ›Reveal solutionSolution

   Applying $\sum m_i c_i (T-T_i)=0$ with effective heat capacities in the ratio $mc:4mc:1.5mc$ gives $T = \dfrac{260}{6.5} = 40^\circ$C.

   Let $c$ be the specific heat capacity of liquid A. Then $c_B = 2c$ and $c_{\text{vessel}} = 0.3c$.

   Effective heat capacities:

   • Liquid A: $m\cdot c = mc$ at $30^\circ$C
   • Liquid B: $2m\cdot 2c = 4mc$ at $50^\circ$C
   • Vessel: $5m\cdot 0.3c = 1.5mc$ at $20^\circ$C …

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7. TG EAPCET 2026Set ap-2026-05-05-FN1 markMCQ
   Q.A hot liquid P of specific heat capacity S is mixed with a cold liquid Q of mass 40 g and specific heat capacity 1.5S. If the fall in temperature of liquid P is 3 times the rise in temperature of liquid Q, then the mass of the liquid P is (A) 40 g (B) 20 g (C) 30 g (D) 50 g

   ›Reveal solutionSolution

   The key is to equate the heat lost by P to the heat gained by Q, using the given ratio of temperature changes. The mass of P comes out to be 20 g.

   The problem is a classic calorimetry question. When two substances at different temperatures are mixed, heat flows from the hotter to the colder until thermal equilibrium is reached. The fundamental principle is that the heat lost by the hot substance equals the heat gained by the cold substance, assuming no heat is lost to the surroundings.

   Here, we are told that the fall in temperature of liquid P is three times the rise in temperature of liquid Q. This ratio is the crucial link that lets us find the unknown mass without needing the actual temperatures.

   Let’s work through it step by step.

   1. Define the variables.

      Let the mass of liquid P be $m_P$ grams (we’ll keep units in grams and calories for convenience).

      Its specific heat capacity is $S$.

      Let the fall in its temperature be $\Delta T_P$.

      For liquid Q, mass $m_Q = 40$ g, specific heat capacity $1.5S$, and rise in temperature $\Delta T_Q$.

   2. Write the heat equations.

      Heat lost by P:

   $Q_{\text{lost}} = m_P \cdot S \cdot \Delta T_P$

   Heat gained by Q:

   $Q_{\text{gained}} = 40 \cdot (1.5S) \cdot \Delta T_Q = 60 S \cdot \Delta T_Q$

   3. Apply the principle of calorimetry. Heat lost = Heat gained:

   $m_P \cdot S \cdot \Delta T_P = 60 S \cdot \Delta T_Q$

   Notice that $S$ cancels out from both sides (as long as $S \neq 0$, which is true for any real liquid).

   $m_P \cdot \Delta T_P = 60 \cdot \Delta T_Q$

   4. Use the given ratio of temperature changes. The problem states: “the fall in temperature of liquid P is 3 times the rise in temperature of liquid Q.” That is:

   $\Delta T_P = 3 \cdot \Delta T_Q$

   5. Substitute and solve. …

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8. TG EAPCET 2026Set ap-2026-05-05-FN1 markMCQ
   Q.If an electric kettle of power $1.5~\text{kW}$ connected to $220~\text{V}$ supply increases the temperature of water of mass $1600~\text{g}$ from $20~^\circ\text{C}$ to $95~^\circ\text{C}$ in $7$ minutes, then the efficiency of the kettle is (Mechanical equivalent of heat $= 4.2~\text{J cal}^{-1}$) (A) $60\%$ (B) $75\%$ (C) $80\%$ (D) $90\%$

   ›Reveal solutionSolution

   The kettle’s efficiency is the ratio of the heat actually gained by the water to the electrical energy consumed. After computing both, the efficiency comes out to 80%.

   The core idea here is simple: efficiency compares useful output to total input. The kettle’s job is to heat water, so the useful output is the heat absorbed by the water. The total input is the electrical energy drawn from the supply over the 7 minutes. The ratio, expressed as a percentage, is the efficiency.

   We need to be careful with units — power is in kW, time in minutes, mass in grams, and the mechanical equivalent of heat tells us how many joules correspond to one calorie. Let’s work through it step by step.

   1. Find the heat required to warm the water. The water’s mass is $m = 1600~\text{g}$. The temperature rise is $\Delta T = 95 - 20 = 75~^\circ\text{C}$. The specific heat capacity of water is $1~\text{cal g}^{-1}~^\circ\text{C}^{-1}$ (a standard value). So the heat needed in calories is:

   $Q_\text{cal} = m \, c \, \Delta T = 1600 \times 1 \times 75 = 120000~\text{cal}.$

   2. Convert that heat into joules. The mechanical equivalent of heat is $4.2~\text{J cal}^{-1}$, meaning 1 calorie = 4.2 joules.

   $Q_\text{J} = 120000 \times 4.2 = 504000~\text{J}.$

   This is the useful energy output.

   3. Calculate the electrical energy consumed by the kettle. Power $P = 1.5~\text{kW} = 1500~\text{W}$ (since 1 kW = 1000 W). Time $t = 7~\text{minutes} = 7 \times 60 = 420~\text{s}$. Electrical energy input is:

   $E_\text{in} = P \times t = 1500 \times 420 = 630000~\text{J}.$

   4. Compute the efficiency. Efficiency $\eta$ is (useful output / total input) × 100%: …

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9. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.The length of a metal rod is 20 cm and its area of cross-section is $4~\text{cm}^2$. If one end of the rod is kept at a temperature of $100~^\circ\text{C}$ and the other end is kept in ice at $0~^\circ\text{C}$, then the mass of the ice melted in 7 minutes is (Thermal conductivity of the metal $= 90~\text{Wm}^{-1}\text{K}^{-1}$ and latent heat of fusion of ice $= 336 \times 10^3~\text{Jkg}^{-1}$) (A) $90~\text{g}$ (B) $67.5~\text{g}$ (C) $22.5~\text{g}$ (D) $45~\text{g}$

   ›Reveal solutionSolution

   The problem is a steady-state heat conduction scenario where the heat conducted through the rod melts ice. Using Fourier’s law, the heat transferred in 7 minutes is calculated, then divided by the latent heat to find the mass of ice melted. The result is 0.0225 kg = 22.5 g, so the correct option is (C).

   The key concept here is steady-state heat conduction through a uniform rod. When the two ends are held at fixed temperatures, heat flows at a constant rate from the hot end to the cold end. That heat is entirely used to melt the ice at the cold end. The rate of heat transfer is given by Fourier’s law, and the total heat over a given time is simply that rate multiplied by time. Then, using the latent heat of fusion, we find how much ice that heat can melt.

   Let’s work through it step by step.

   1. Identify the given data and convert units where necessary.

      • Length of rod, $L = 20 \, \text{cm} = 0.20 \, \text{m}$
      • Cross-sectional area, $A = 4 \, \text{cm}^2 = 4 \times 10^{-4} \, \text{m}^2$
      • Temperature difference, $\Delta T = 100^\circ\text{C} - 0^\circ\text{C} = 100 \, \text{K}$ (since a change of 1°C equals 1 K)
      • Thermal conductivity, $k = 90 \, \text{W m}^{-1} \text{K}^{-1}$
      • Time, $t = 7 \, \text{minutes} = 7 \times 60 = 420 \, \text{s}$
      • Latent heat of fusion of ice, $L_f = 336 \times 10^3 \, \text{J kg}^{-1}$

   2. Apply Fourier’s law for steady-state heat conduction.

      The rate of heat flow (power) through the rod is:

   $P = \frac{k A \Delta T}{L}$

   Substitute the values:

   $P = \frac{90 \times (4 \times 10^{-4}) \times 100}{0.20}$

   First, compute numerator: $90 \times 4 \times 10^{-4} \times 100 = 90 \times 4 \times 10^{-2} = 360 \times 10^{-2} = 3.6$

   Then divide by 0.20: $3.6 / 0.20 = 18$

   So, $P = 18 \, \text{W}$ (joules per second).

   3. Calculate the total heat transferred in 7 minutes.

   $Q = P \times t = 18 \times 420 = 7560 \, \text{J}$

   4. Relate this heat to the mass of ice melted. …

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10. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
    Q.A rectangular ice box of total surface area of $1000~\mathrm{cm}^2$ initially contains $1.5~\mathrm{kg}$ of ice at $0~^\circ\mathrm{C}$. If the thickness of the walls of the box is $2~\mathrm{mm}$ and the temperature outside the box is $42~^\circ\mathrm{C}$, then the mass of the ice remaining in the box after $160$ minutes is (Thermal conductivity of the material of the box $= 10^{-2}~\mathrm{W\,m^{-1}K^{-1}}$ and latent heat of the fusion of ice $= 336 \times 10^3~\mathrm{J\,kg^{-1}}$) (A) $0.6~\mathrm{kg}$ (B) $0.9~\mathrm{kg}$ (C) $0.8~\mathrm{kg}$ (D) $0.7~\mathrm{kg}$

    ›Reveal solutionSolution

    The heat conducted through the box walls melts the ice; using the conduction formula and latent heat, the melted mass is found to be 0.9 kg, so the remaining ice is 0.6 kg. The correct option is (A).

    Concept & Intuition

    The problem is a classic example of heat transfer by conduction driving a phase change. The ice inside stays at 0 °C while melting, and the outside is at 42 °C. Heat flows through the walls at a steady rate given by Fourier’s law. That heat is entirely used to melt the ice (latent heat). So we can compute the total heat conducted over 160 minutes, then find how much ice that melts, and subtract from the initial mass.

    ---

    1. Identify the driving temperature difference The inside is at $0^\circ\mathrm{C}$ (ice melting), outside at $42^\circ\mathrm{C}$.

    $\Delta T = 42 - 0 = 42\ \mathrm{K}$

    2. Convert all units to SI

       • Surface area: $1000\ \mathrm{cm}^2 = 1000 \times 10^{-4} = 0.1\ \mathrm{m}^2$
       • Wall thickness: $2\ \mathrm{mm} = 2 \times 10^{-3}\ \mathrm{m}$
       • Time: $160\ \mathrm{min} = 160 \times 60 = 9600\ \mathrm{s}$
       • Thermal conductivity: $k = 10^{-2}\ \mathrm{W\,m^{-1}K^{-1}}$
       • Latent heat: $L = 336 \times 10^3\ \mathrm{J\,kg^{-1}}$

    3. Apply Fourier’s law of heat conduction

       The rate of heat flow through the walls is

    $\frac{dQ}{dt} = \frac{k A \Delta T}{d}$

    where $d$ is thickness. Substitute:

    $\frac{dQ}{dt} = \frac{(10^{-2})(0.1)(42)}{2 \times 10^{-3}} = \frac{0.042}{0.002} = 21\ \mathrm{W}$

    So 21 J of heat flows in every second.

    4. Total heat conducted in 160 minutes

    $Q = \left(\frac{dQ}{dt}\right) \times t = 21 \times 9600 = 201600\ \mathrm{J}$

    5. Mass of ice melted by this heat The heat required to melt mass $m$ of ice at 0 °C is $Q = m L$. Thus $$$$

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11. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
    Q.A rectangular ice box of total surface area of $1000~\mathrm{cm}^2$ initially contains $1.5~\mathrm{kg}$ of ice at $0~^\circ\mathrm{C}$. If the thickness of the walls of the box is $2~\mathrm{mm}$ and the temperature outside the box is $42~^\circ\mathrm{C}$, then the mass of the ice remaining in the box after $160$ minutes is (Thermal conductivity of the material of the box $= 10^{-2}~\mathrm{Wm}^{-1}\mathrm{K}^{-1}$ and latent heat of the fusion of ice $= 336 \times 10^3~\mathrm{Jkg}^{-1}$) (A) $0.8~\mathrm{kg}$ (B) $0.9~\mathrm{kg}$ (C) $0.6~\mathrm{kg}$ (D) $0.7~\mathrm{kg}$

    ›Reveal solutionSolution

    The problem is solved by equating the heat conducted through the box walls to the heat absorbed by the ice for melting. Using the conduction formula and latent heat, the mass melted is 0.6 kg, leaving 0.9 kg. The correct option is (B).

    Concept and Intuition

    This is a classic heat transfer problem where thermal energy flows from the warm outside into the cold interior. The ice at 0°C absorbs this energy and melts, but the box walls resist the flow. The key is that the rate of heat conduction depends on the temperature difference, wall area, thickness, and material conductivity. Over a given time, the total heat conducted equals the heat needed to melt a certain mass of ice. We don’t need to worry about the ice temperature changing—it stays at 0°C until all ice melts—so the calculation is straightforward.

    Step-by-step solution

    1. Identify the known quantities

       • Total surface area of the box: $A = 1000~\mathrm{cm}^2 = 0.1~\mathrm{m}^2$ (since $1~\mathrm{m}^2 = 10^4~\mathrm{cm}^2$).
       • Wall thickness: $d = 2~\mathrm{mm} = 0.002~\mathrm{m}$.
       • Outside temperature: $T_{\text{out}} = 42^\circ\mathrm{C}$.
       • Inside temperature (ice at 0°C): $T_{\text{in}} = 0^\circ\mathrm{C}$.
       • Temperature difference: $\Delta T = 42 - 0 = 42~\mathrm{K}$.
       • Thermal conductivity: $k = 10^{-2}~\mathrm{W\,m^{-1}K^{-1}}$.
       • Time: $t = 160~\text{minutes} = 160 \times 60 = 9600~\mathrm{s}$.
       • Latent heat of fusion: $L = 336 \times 10^3~\mathrm{J\,kg^{-1}}$.
       • Initial mass of ice: $m_i = 1.5~\mathrm{kg}$.

    2. Calculate the rate of heat conduction

       Fourier’s law for heat conduction through a slab is:

    $\frac{Q}{t} = \frac{k A \Delta T}{d}$

    Substitute the values:

    $\frac{Q}{t} = \frac{(10^{-2})(0.1)(42)}{0.002}$

    First, numerator: $10^{-2} \times 0.1 = 10^{-3}$, times 42 gives $4.2 \times 10^{-2}$.

    Divide by 0.002: $4.2 \times 10^{-2} / 0.002 = 21$.

    So the heat flow rate is $21~\mathrm{W}$ (joules per second). …

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12. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
    Q.If a radioactive substance decays 10% in every 16 hours, then the percentage of the radioactive substance that remains after 2 days is (A) 72.9 (B) 82.2 (C) 18.8 (D) 27.1

    ›Reveal solutionSolution

    This is a repeated percentage decay problem: after each 16‑hour period, 90% remains. Two days = 48 hours = three 16‑hour periods, so the fraction left is $0.9^3 = 0.729$, i.e. 72.9%, which corresponds to option (A).

    The key idea is that “decays 10%” means “retains 90%” each fixed time interval. Because the decay happens in discrete, equal steps, we can simply multiply the remaining fraction repeatedly.

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    1. Interpret the decay rate

       “Decays 10%” means that after 16 hours, the substance loses 10% of its current amount, so it keeps $100\% - 10\% = 90\%$.

       As a decimal, the retention factor per 16 hours is $0.9$.

    2. Find how many 16‑hour periods fit into 2 days

       2 days = 48 hours.

       Number of periods = $\frac{48}{16} = 3$.

    3. Apply the decay repeatedly

       Starting with an initial amount $A_0$, after one period: $A_1 = A_0 \times 0.9$.

       After two periods: $A_2 = A_1 \times 0.9 = A_0 \times 0.9^2$.

       After three periods: $A_3 = A_0 \times 0.9^3$.

    4. Compute the final percentage

       $0.9^3 = 0.9 \times 0.9 \times 0.9 = 0.81 \times 0.9 = 0.729$ …

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