Q.A $10\ \text{kW}$ drilling machine is used to drill a bore in a small aluminium block of mass $8.0\ \text{kg}$. How much is the rise in temperature of the block in $2.5$ minutes, assuming $50\%$ of power is used up in heating the machine itself or lost to the surroundings. Specific heat of aluminium $= 0.91\ \text{J g}^{-1}\ \text{K}^{-1}$.

Showing the 12 most recent of 37 on this concept.

1. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.An earthen pitcher containing 9.5 kg of water loses one gram of water per minute due to evaporation. If water equivalent of pitcher is 0.5 kg. The time required to cool the water in pitcher from $30°C$ to $28°C$ is (Neglect radiation effect and Take latent heat of vaporization $500\ cal\,g^{-1}$) (A) 30 min (B) 60 min (C) 40 min (D) 20 min

   ›Reveal solutionSolution

   Tests using evaporative cooling (latent heat) to find how long it takes to cool a given water equivalent by a given temperature drop. Answer: 40 min.

   Concept and Intuition

   Evaporation is a cooling mechanism: the fastest molecules escape the liquid surface carrying away the latent heat of vaporization, which comes at the expense of the remaining liquid's thermal energy — hence its temperature drops. The problem tells us to neglect radiative losses, so the only mechanism cooling the pitcher's contents is evaporation, letting us equate the heat required to cool the water equivalent by $2°C$ to the heat carried away by the evaporated mass.

   Step-by-Step Solution

   1. Water equivalent of the whole system (water + pitcher) $=9.5+0.5=10$ kg $=10000$ g.
   2. Heat that must leave the system to cool it by $\Delta T=30-28=2°C$: $Q=mc\Delta T=10000\times1\times2=20000$ cal (using $c=1\ cal\,g^{-1}°C^{-1}$ for the water-equivalent basis). …

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2. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.Heat energy released by water of mass $3\ kg$ when it is cooled by $20^\circ C$ is (specific heat capacity of water $= 4200\ J\,kg^{-1}K^{-1}$) (A) 252000 J (B) 420000 J (C) 52000 J (D) 25200 J

   ›Reveal solutionSolution

   This is a direct application of $Q=mc\Delta T$. The answer is 252000 J.

   Concept and Intuition

   The heat energy released or absorbed by a substance when its temperature changes (without a phase change) is given by $Q = mc\Delta T$, where $m$ is mass, $c$ is specific heat capacity, and $\Delta T$ is the temperature change. Here water cools down, so this heat is released to the surroundings.

   Step-by-Step Solution

   1. Given: $m = 3\ kg$, $c = 4200\ J\,kg^{-1}K^{-1}$, $\Delta T = 20^\circ C = 20\ K$ (a temperature difference is the same in Celsius and Kelvin).
   2. Compute: $Q = mc\Delta T = 3 \times 4200 \times 20$.
   3. $3 \times 4200 = 12600$; then $12600 \times 20 = 252000\ J$. …

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3. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.A beaker contains 200 gm of water. The heat capacity of the beaker is equal to that of 20 gm of water. The initial temperature of water in the beaker is $20^0C$. If 440 gm of hot water at $92^0C$ is poured in it, the final temperature (neglecting radiation loss) will be nearest to (A) $58^0C$ (B) $68^0C$ (C) $78^0C$ (D) $73^0C$

   ›Reveal solutionSolution

   A calorimetry problem where the beaker's own heat capacity must be folded in as "equivalent water mass." Final equilibrium temperature works out to $68^0C$.

   Concept and Intuition

   When mixing two masses of water inside a container, the container itself absorbs some heat too. Rather than tracking the beaker's material and specific heat separately, we're given its heat capacity directly in terms of an equivalent mass of water (20 g) — this is the standard "water equivalent" trick: any object with heat capacity equal to $m_{eq}$ grams of water behaves, thermally, exactly like that much extra water. So the initial cold system (beaker + water) behaves like $200+20=220$ g of water at $20^0C$.

   Step-by-Step Solution

   1. Water-equivalent of cold side: $200\ g\ (\text{water}) + 20\ g\ (\text{beaker's equivalent}) = 220\ g$ at $20^0C$.
   2. Hot water added: $440\ g$ at $92^0C$.
   3. By conservation of energy (no radiation loss), heat lost by hot water = heat gained by cold side: $m_{hot}\,c\,(92-T) = m_{cold,eq}\,c\,(T-20)$ …

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4. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.A geyser heats water flowing at the rate of 3.0 liters per minute from $27^{0}C$ to $77^{0}C$. If the geyser operates on a gas burner and if its heat of combustion is $4.0\times10^4\ Jg^{-1}$, the rate of combustion of the fuel per minute is (A) $15.75\times10^{-3}$ g (B) 15.75 g (C) 252 g (D) $252\times10^{-3}$ g

   ›Reveal solutionSolution

   Equate the heat absorbed by the flowing water per minute to the heat released by burning fuel; solving for fuel mass gives 15.75 g per minute.

   Concept and Intuition

   This is a straightforward calorimetry + combustion-energy balance: whatever heat the burner supplies (mass of fuel burnt × its heat of combustion) must equal the heat gained by the water flowing through the geyser (mass of water per minute × specific heat × temperature rise), assuming no losses.

   Step-by-Step Solution

   1. Water flow rate: 3.0 L/min $= 3000$ g/min (density of water $\approx 1$ g/mL).
   2. Temperature rise: $\Delta T = 77-27 = 50^{0}$C.
   3. Heat required per minute: $Q = mc\Delta T = 3000\ \text{g}\times4.2\ \text{J g}^{-1}{}^{0}\text{C}^{-1}\times50\ ^{0}\text{C} = 6.3\times10^5$ J. …

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5. AP EAPCET 2026Set ap-2026-05-19-AN1 markMCQ
   Q.A metal block of mass 10 kg freely falls from a height 1m and rebounds back to 58 cm. If the loss in energy is completely utilized by the block, then the rise in temperature of the block is (Acceleration due to gravity = 10 $\mathrm{ms^{-2}}$. Specific heat of the material is 0.05 cal $\mathrm{g^{-1}\ ^{\circ}C^{-1}}$) (A) $2^{\circ}\mathrm{C}$ (B) $5^{\circ}\mathrm{C}$ (C) $0.2^{\circ}\mathrm{C}$ (D) $0.02^{\circ}\mathrm{C}$

   ›Reveal solutionSolution

   The 42 J of mechanical energy lost on the bounce converts entirely to heat, raising the block's temperature by $0.02^\circ$C.

   Concept and Intuition

   When a falling block rebounds to a lower height than it fell from, mechanical energy has been "lost" — but energy is still conserved overall; it has been converted into heat within the block (assuming, as stated, that all the lost energy goes into heating the block).

   Step-by-Step Solution

   1. Energy just before impact (from falling height $h_1=1$ m): $E_1 = mgh_1 = 10\times10\times1 = 100\ \text{J}$.
   2. Energy just after rebound (rising to $h_2=0.58$ m): $E_2 = mgh_2 = 10\times10\times0.58 = 58\ \text{J}$.
   3. Energy lost (converted to heat): $\Delta E = E_1-E_2 = 100-58 = 42\ \text{J}$.
   4. Convert to calories (using $1\ \text{cal}=4.2\ \text{J}$): $Q = 42/4.2 = 10\ \text{cal}$. …

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6. AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQ
   Q.Certain amount of heat is given to 100 gm of copper to increase its temperature by $21^0C$. If the same amount of heat is given to 50 gm of water, then the raise in its temperature is (specific heat capacity of copper $= 400\,\mathrm{J\,kg^{-1}\,K^{-1}}$ and that of water $= 4200\,\mathrm{J\,kg^{-1}\,K^{-1}}$) (A) $4^0C$ (B) $5.25^0C$ (C) $8^0C$ (D) $6^0C$

   ›Reveal solutionSolution

   Compute the heat absorbed by the copper first, then use the same heat quantity with water's specific heat to find its temperature rise: $4^\circ$C.

   Concept and Intuition

   The heat required to raise a mass $m$ of a substance by $\Delta T$ is $Q=mc\Delta T$, where $c$ is the specific heat capacity. Since the same amount of heat $Q$ is given to both substances (just at different times), we can first calculate $Q$ from the copper data, then use it to solve for water's temperature rise.

   Step-by-Step Solution

   1. For copper: $m_{Cu}=0.1\,\mathrm{kg}$, $c_{Cu}=400\,\mathrm{J\,kg^{-1}K^{-1}}$, $\Delta T_{Cu}=21\,\mathrm K$.
   2. $Q = m_{Cu}c_{Cu}\Delta T_{Cu} = 0.1\times400\times21 = 840\,\mathrm J$.
   3. For water: $m_w=0.05\,\mathrm{kg}$, $c_w=4200\,\mathrm{J\,kg^{-1}K^{-1}}$. …

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7. AP EAPCET 2026Set ap-2026-05-20-AN1 markMCQ
   Q.Two liquids A and B are at temparatures 40°C and 20°C respectively. When equal masses of these liquids are mixed, the temperature of the mixture is found to be 35°C. The ratio of specific heats of A and B is (A) 1 : 3 (B) 3 : 1 (C) 2 : 1 (D) 1 : 2

   ›Reveal solutionSolution

   Applying calorimetry (heat lost by the hotter liquid = heat gained by the cooler one) to the equal-mass mixture gives specific-heat ratio $c_A:c_B = 3:1$. Answer: (B).

   Concept and Intuition

   When two liquids of equal mass but different temperatures and specific heats are mixed (with no heat lost to surroundings), the heat given up by the hotter liquid as it cools to the final temperature exactly equals the heat absorbed by the colder liquid as it warms up. Since both masses are equal, they cancel out of the equation, leaving a direct relation between the specific heats and the temperature changes each liquid undergoes.

   Step-by-Step Solution

   1. Liquid A: temperature drops from $40^\circ C$ to $35^\circ C$, i.e., $\Delta T_A = 5^\circ C$.
   2. Liquid B: temperature rises from $20^\circ C$ to $35^\circ C$, i.e., $\Delta T_B = 15^\circ C$. …

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8. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
   Q.An electric kettle takes 4 A current at 220 V. If the entire electric energy is converted into heat energy, then the time (in minutes) taken to increase the temperature of 1 kg of water from $34^\circ\mathrm{C}$ to $100^\circ\mathrm{C}$ is (A) 7.50 (B) 4.50 (C) 5.25 (D) 6.25

   ›Reveal solutionSolution

   Electrical energy converted entirely to heat raises 1 kg of water by 66°C; equating electrical energy to the required heat gives a time of 5.25 minutes.

   Concept and Intuition

   When all electrical energy is converted to heat (no losses), the energy supplied by the kettle equals the thermal energy needed to raise the water's temperature: $VIt = mc\Delta T$. This is simply energy conservation applied to Joule heating.

   Step-by-Step Solution

   1. Power of the kettle: $P = VI = 220\times4 = 880\ \mathrm{W}$.
   2. Heat required to raise 1 kg of water from $34^\circ\mathrm{C}$ to $100^\circ\mathrm{C}$: $Q = mc\Delta T = (1)(4200)(100-34) = 4200\times66 = 277200\ \mathrm{J}$.
   3. Time: $t = Q/P = 277200/880 = 315\ \mathrm{s}$. …

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9. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.Water of mass 5 kg in a closed vessel is at a temperature of 20 °C. If the temperature of the water when heated for a time of 10 minutes becomes 30 °C, then the increase in the internal energy of the water is (Specific heat capacity of water = 4200 J kg$^{-1}$ K$^{-1}$) (A) 100 kJ (B) 420 kJ (C) 510 kJ (D) 210 kJ

   ›Reveal solutionSolution

   This tests the first law of thermodynamics for a rigid, closed container — since no work is done, the heat added equals the rise in internal energy. The answer is 210 kJ.

   Concept and Intuition

   Internal energy is the total kinetic + potential energy of a system's molecules. The first law states $Q = \Delta U + W$, where $W$ is work done BY the system. In a closed, rigid vessel, the water's volume cannot change, so it cannot push against anything or expand — hence $W = 0$. Every joule of heat supplied therefore shows up entirely as a rise in internal energy; none is 'spent' doing mechanical work.

   Step-by-Step Solution

   1. First law: $Q = \Delta U + W$.
   2. The vessel is closed (rigid, fixed volume) ⟹ no expansion work is done: $W = 0$.
   3. So $\Delta U = Q$, the full heat supplied.
   4. Heat supplied: $Q = mc\Delta T$, with $m = 5$ kg, $c = 4200\ \mathrm{J\,kg^{-1}K^{-1}}$, $\Delta T = 30 - 20 = 10$ K. …

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10. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
    Q.Steam at $100\ ^\circ\mathrm{C}$ is passed into 114 g of water at $30\ ^\circ\mathrm{C}$. The mass of water present in the mixture when the temperature of the water becomes $70\ ^\circ\mathrm{C}$ is (Latent heat of steam $= 540\ \mathrm{cal\,g^{-1}}$; specific heat capacity of water $= 1\ \mathrm{cal\,g^{-1}\,^\circ C^{-1}}$) (A) 122 g (B) 132 g (C) 142 g (D) 152 g

    ›Reveal solutionSolution

    Calorimetry problem: steam condenses and cools, giving up heat that warms the original water; find how much steam condenses and add it to the original mass. Answer: 122 g.

    Concept and Intuition

    When steam is passed into cooler water, it first condenses (releasing latent heat) and then the resulting hot water cools further, releasing sensible heat, until thermal equilibrium is reached with the original water (which simultaneously warms up). Conservation of heat energy (heat lost by steam = heat gained by water) lets us find the mass of steam that condensed; this condensed mass adds to the original water mass.

    Step-by-Step Solution

    1. Let $m$ grams of steam condense. Heat released by steam = (heat released condensing at 100°C) + (heat released cooling from 100°C to 70°C) $= m(540) + m(1)(100-70) = 570m$ cal.
    2. Heat absorbed by the original 114 g of water warming from 30°C to 70°C: $Q = 114 \times 1 \times (70-30) = 114\times40 = 4560$ cal. …

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11. AP EAPCET 2025Set ap-2025-05-20-FN1 markMCQ
    Q.500 g of water at a temperature of 73 °C is mixed with 45 g of steam at a temperature of 100 °C. The ratio of the initial mass of steam and the mass of steam at equilibrium of the mixture is (A) 9 : 4 (B) 3 : 1 (C) 3 : 2 (D) 5 : 4

    ›Reveal solutionSolution

    Since 45 g of steam carries far more latent heat than needed to bring 500 g of water from 73°C to 100°C, the mixture settles at 100°C with some steam still uncondensed; the initial-to-final steam mass ratio works out to 9:4.

    Concept and Intuition

    When steam is mixed with water that is below 100°C, the steam first condenses (releasing its large latent heat) to warm the water. If there's more steam than needed to bring the water exactly to 100°C, the mixture equilibrates AT 100°C, with the "leftover" steam remaining as vapor (a boiling water + steam equilibrium) rather than the temperature rising further, since water can't exceed 100°C at this pressure while steam is still present.

    Step-by-Step Solution

    1. Heat needed to raise 500 g of water from 73°C to 100°C:

    $Q = mc\Delta T = 500 \times 1 \times (100-73) = 500 \times 27 = 13500\ \text{cal}$

    2. Suppose a mass $m$ (in grams) of the 45 g steam condenses to supply this heat via its latent heat $L = 540\ \text{cal/g}$:

    $mL = 13500 \implies m = \frac{13500}{540} = 25\ \text{g}$

    3. Since $25\ \text{g} < 45\ \text{g}$, this is self-consistent — not all the steam is needed, so the equilibrium temperature is indeed 100°C with steam remaining. …

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12. AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
    Q.A small quantity of water of mass 'm' at temperature $\theta\ ^\circ\mathrm{C}$ is mixed with a large mass 'M' of ice which is at its melting point. If 's' is specific heat capacity of water and 'L' is the Latent heat of fusion of ice, then the mass of ice melted is (A) $\dfrac{ML}{ms\theta}$ (B) $\dfrac{ms\theta}{ML}$ (C) $\dfrac{Ms\theta}{L}$ (D) $\dfrac{ms\theta}{L}$

    ›Reveal solutionSolution

    Tests the principle of calorimetry (heat lost = heat gained) applied to water cooling down while melting a large ice reservoir. Answer: $\dfrac{ms\theta}{L}$.

    Concept and Intuition

    Since the mass of ice $M$ is large, the ice-water mixture stays at $0^\circ$C throughout (there's always ice left to absorb heat without the mixture's temperature rising). All the heat given up by the small quantity of warm water as it cools to $0^\circ$C goes into melting some of the ice — none is left over to raise the temperature of anything, since excess ice remains solid at the end.

    Step-by-Step Solution

    1. Heat lost by the water of mass $m$ cooling from $\theta^\circ$C to $0^\circ$C:

    $Q_{lost} = m s \theta$

    2. This heat is absorbed by a mass $m'$ of ice melting at its melting point (latent heat $L$, no temperature change involved since melting is at constant temperature): $Q_{gained} = m' L$ …

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