Q.What amount of heat must be supplied to $2.0 \times 10^{-2}\ \text{kg}$ of nitrogen (at room temperature) to raise its temperature by $45\,^{\circ}\text{C}$ at constant pressure? (Molecular mass of $\text{N}_2 = 28$; $R = 8.3\ \text{J mol}^{-1}\ \text{K}^{-1}$.)

Showing the 12 most recent of 36 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.5 moles of monoatomic gas and one mole of rigid diatomic gas are mixed. The internal energy (in kJ) of the gaseous mixture at a temperature of 127 $^\circ$C is (Universal gas constant = 8.31 J mol$^{-1}$ K$^{-1}$) (A) 66.48 (B) 49.86 (C) 16.62 (D) 33.24

   ›Reveal solutionSolution

   Total internal energy $U=\left(\tfrac{3}{2}n_1+\tfrac{5}{2}n_2\right)RT=10RT=33.24\,\mathrm{kJ}$.

   For $n_1=5$ moles monoatomic gas, $C_V=\tfrac{3}{2}R$; for $n_2=1$ mole rigid diatomic gas, $C_V=\tfrac{5}{2}R$. Temperature $T=127\,^\circ\mathrm{C}=400\,\mathrm{K}$.

   $U=\left(n_1\cdot\tfrac{3}{2}R+n_2\cdot\tfrac{5}{2}R\right)T=\left(\tfrac{15}{2}+\tfrac{5}{2}\right)RT=10\,RT$

   $U=10\times8.31\times400=33240\,\mathrm{J}=33.24\,\mathrm{kJ}$

   ✓Final answer

   $U=33.24\,\mathrm{kJ}$ — option (D).

2. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.If $Q_1$ is the heat required to convert 2 g of ice at a temperature of $0\,^\circ \mathrm{C}$ to water at a temperature of $40\,^\circ \mathrm{C}$ and $Q_2$ is the heat required to convert 4 g of ice at a temperature of $0\,^\circ \mathrm{C}$ to water at a temperature of $80\,^\circ \mathrm{C}$, then $Q_1 : Q_2 =$ (Latent heat of fusion of ice $= 80\,\mathrm{cal}\,\mathrm{g}^{-1}$ and specific heat capacity of water $= 1\,\mathrm{cal}\,\mathrm{g}^{-1}\,^\circ \mathrm{C}^{-1}$) (A) $3 : 8$ (B) $1 : 2$ (C) $1 : 4$ (D) $3 : 4$

   ›Reveal solutionSolution

   The ratio $Q_1 : Q_2$ is found by calculating the total heat (latent + sensible) for each case and simplifying. The result is $3 : 8$.

   The problem asks for the ratio of two heats, each involving two stages: melting ice at $0^\circ \mathrm{C}$ into water at $0^\circ \mathrm{C}$ (latent heat), then heating that water to a final temperature (sensible heat). The key is to treat each stage separately and sum them.

   Latent heat depends only on mass, not on temperature change. Sensible heat depends on mass, specific heat, and the temperature rise. So for each case, we compute $Q = m L_f + m c \Delta T$, then take the ratio.

   1. For $Q_1$ (2 g ice to $40^\circ \mathrm{C}$ water):

      Mass $m_1 = 2\,\mathrm{g}$.

      Latent heat: $Q_{\text{latent},1} = m_1 L_f = 2 \times 80 = 160\,\mathrm{cal}$.

      Sensible heat (heating water from $0^\circ \mathrm{C}$ to $40^\circ \mathrm{C}$): $Q_{\text{sensible},1} = m_1 c \Delta T_1 = 2 \times 1 \times 40 = 80\,\mathrm{cal}$.

      Total: $Q_1 = 160 + 80 = 240\,\mathrm{cal}$.

   2. For $Q_2$ (4 g ice to $80^\circ \mathrm{C}$ water):

      Mass $m_2 = 4\,\mathrm{g}$.

      Latent heat: $Q_{\text{latent},2} = 4 \times 80 = 320\,\mathrm{cal}$.

      Sensible heat (heating from $0^\circ \mathrm{C}$ to $80^\circ \mathrm{C}$): $Q_{\text{sensible},2} = 4 \times 1 \times 80 = 320\,\mathrm{cal}$.

      Total: $Q_2 = 320 + 320 = 640\,\mathrm{cal}$.

   3. Find the ratio:

      $\displaystyle \frac{Q_1}{Q_2} = \frac{240}{640} = \frac{3}{8}$.

      So $Q_1 : Q_2 = 3 : 8$.

   Watch out

   A common mistake is to forget the latent heat part and only compare the sensible heats, which would give $1:4$ — option (C). Always account for the phase change first.

   Tip

   Notice that both masses and temperature changes are in simple multiples. You could write $Q_1 = m_1(L_f + c\Delta T_1)$ and $Q_2 = m_2(L_f + c\Delta T_2)$, then take the ratio directly: $\frac{2(80+40)}{4(80+80)} = \frac{2 \times 120}{4 \times 160} = \frac{240}{640} = \frac{3}{8}$.

   ✓Final answer

   The ratio is $\boxed{3:8}$, which corresponds to option (A).

3. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.Ice of mass 80 g at a temperature of $-10\,^\circ\mathrm{C}$ is mixed with water of mass 100 g at a temperature of $20\,^\circ\mathrm{C}$. The ratio of the masses of ice and water in the mixture in equilibrium is (Latent heat of fusion of ice $= 80\,\mathrm{cal\,g}^{-1}$, specific heat capacities of water and ice are $1\,\mathrm{cal\,g}^{-1}\mathrm{^\circ C}^{-1}$ and $0.5\,\mathrm{cal\,g}^{-1}\mathrm{^\circ C}^{-1}$ respectively) (A) $4:5$ (B) $1:2$ (C) $2:3$ (D) $3:4$

   ›Reveal solutionSolution

   The key idea is to check whether all the ice melts or not by comparing the heat needed to warm and melt the ice with the heat the water can supply. The final equilibrium mixture has ice and water in the ratio 1:2, so the correct option is (B).

   ---

   Concept and Intuition

   When ice and water at different temperatures are mixed, heat flows from the warmer water to the colder ice. The ice first warms to 0 °C, then may melt if enough heat remains. The water cools to 0 °C. The final state depends on whether the water’s available heat is enough to melt all the ice. We calculate the heat required to bring the ice to 0 °C and then melt it, and compare it with the heat the water can give up by cooling to 0 °C. If the water’s heat is insufficient, only part of the ice melts, and the mixture stays at 0 °C with both ice and water present.

   ---

   Step-by-step solution

   1. Heat needed to warm the ice from –10 °C to 0 °C Mass of ice $m_i = 80\ \text{g}$, specific heat of ice $c_i = 0.5\ \text{cal g}^{-1}\text{°C}^{-1}$.

   $Q_1 = m_i c_i \Delta T = 80 \times 0.5 \times (0 - (-10)) = 80 \times 0.5 \times 10 = 400\ \text{cal}.$

   2. Heat needed to melt all the ice at 0 °C Latent heat of fusion $L = 80\ \text{cal g}^{-1}$.

   $Q_2 = m_i L = 80 \times 80 = 6400\ \text{cal}.$

   Total heat required to turn all ice into water at 0 °C:

   $Q_{\text{ice, total}} = Q_1 + Q_2 = 400 + 6400 = 6800\ \text{cal}.$

   3. Heat the water can supply by cooling from 20 °C to 0 °C Mass of water $m_w = 100\ \text{g}$, specific heat of water $c_w = 1\ \text{cal g}^{-1}\text{°C}^{-1}$.

   $Q_{\text{water}} = m_w c_w \Delta T = 100 \times 1 \times (20 - 0) = 2000\ \text{cal}.$

   4. Compare the heat available with the heat needed

      The water can supply only 2000 cal, but melting all the ice requires 6800 cal. Clearly, not all ice melts. The mixture will end at 0 °C with some ice remaining.

   5. How much ice actually melts?

      First, the ice must be warmed to 0 °C, which uses 400 cal. The remaining heat from the water is

   $2000 - 400 = 1600\ \text{cal}.$

   This remaining heat goes into melting ice. Mass of ice melted:

   $m_{\text{melted}} = \frac{1600}{80} = 20\ \text{g}.$

   6. Final masses in the mixture
      • Ice remaining: $80 - 20 = 60\ \text{g}$.
      • Water originally: 100 g, plus the 20 g of melted ice gives $100 + 20 = 120\ \text{g}$ of water. Ratio of ice to water (by mass) in the final mixture:

   $\frac{60}{120} = \frac{1}{2}.$

   So the ratio is $1:2$.

   Watch out

   A common mistake is to forget that the ice must first be warmed to 0 °C before any melting can occur. Skipping that step gives a wrong melted mass and a different ratio.

   Tip

   Always check whether the water can supply enough heat to both warm and melt the ice. If not, the final temperature is 0 °C and you only partially melt the ice.

   ✓Final answer

   The correct option is (B).

   ANSWER: B

4. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.The heat required to convert 8 g of ice at a temperature of $-20\,^\circ\mathrm{C}$ to steam at $100\,^\circ\mathrm{C}$ is [Specific heat capacity of ice $= 2100\,\mathrm{J\,kg^{-1}\,K^{-1}}$, specific heat capacity of water $= 4200\,\mathrm{J\,kg^{-1}\,K^{-1}}$, latent heat of fusion of ice $= 336 \times 10^3\,\mathrm{J\,kg^{-1}}$ and latent heat of steam $= 2.268 \times 10^6\,\mathrm{J\,kg^{-1}}$] (A) 5400 cal (B) 5840 cal (C) 5760 cal (D) 5120 cal

   ›Reveal solutionSolution

   Sum the four stages — warming ice, melting, warming water, boiling — to get $24528\ \mathrm{J}$, which is $5840\ \mathrm{cal}$ — option (B).

   ---

   Concept

   Taking $8\ \mathrm{g} = 0.008\ \mathrm{kg}$ from ice at $-20\,^\circ\mathrm{C}$ to steam at $100\,^\circ\mathrm{C}$ needs four heat inputs: raising ice to $0\,^\circ\mathrm{C}$, melting it, raising water to $100\,^\circ\mathrm{C}$, and vaporising it.

   Solution

   $Q_1 = m c_\text{ice}\,\Delta T = 0.008 \times 2100 \times 20 = 336\ \mathrm{J}.$

   $Q_2 = m L_f = 0.008 \times 336\times10^3 = 2688\ \mathrm{J}.$

   $Q_3 = m c_\text{water}\,\Delta T = 0.008 \times 4200 \times 100 = 3360\ \mathrm{J}.$

   $Q_4 = m L_v = 0.008 \times 2.268\times10^6 = 18144\ \mathrm{J}.$

   Total:

   $Q = 336 + 2688 + 3360 + 18144 = 24528\ \mathrm{J}.$

   Converting with $1\ \mathrm{cal} = 4.2\ \mathrm{J}$:

   $Q = \frac{24528}{4.2} = 5840\ \mathrm{cal}.$

   ✓Final answer

   The heat required is $5840\ \mathrm{cal}$ — option (B).

5. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.A gas is suddenly compressed such that its absolute temperature is doubled. If the ratio of the specific heat capacities of the gas is $1.5$, then the percentage decrease in the volume of the gas is (A) 30 (B) 50 (C) 25 (D) 75

   ›Reveal solutionSolution

   For an adiabatic process, $TV^{\gamma-1} = \text{constant}$. With $\gamma = 1.5$ and $T_2 = 2T_1$, the volume ratio is $V_2/V_1 = 1/2\sqrt{2} \approx 0.3536$, so the percentage decrease is about $64.6\%$ — but the given options suggest a different interpretation: the problem likely intends a sudden compression as an adiabatic process where $T \propto V^{1-\gamma}$, leading to a $75\%$ decrease, matching option (D).

   ---

   Concept & Intuition

   The phrase “suddenly compressed” is the key. In thermodynamics, a sudden change means there is no time for heat exchange with the surroundings — the process is adiabatic. For an ideal gas undergoing a reversible adiabatic process, the relation between temperature and volume is

   $T V^{\gamma-1} = \text{constant},$

   where $\gamma = C_p/C_v$ is the ratio of specific heats. Here $\gamma = 1.5$, and the temperature doubles. We can directly find how the volume changes.

   ---

   Step-by-step solution

   1. Write the adiabatic relation For an adiabatic process:

   $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}.$

   Given $T_2 = 2 T_1$ and $\gamma = 1.5$, so $\gamma - 1 = 0.5$.

   2. Substitute and solve for the volume ratio

   $T_1 V_1^{0.5} = (2 T_1) V_2^{0.5}.$

   Cancel $T_1$ (non-zero):

   $V_1^{0.5} = 2 V_2^{0.5}.$

   Square both sides:

   $V_1 = 4 V_2 \quad \Rightarrow \quad \frac{V_2}{V_1} = \frac{1}{4}.$

   3. Interpret the result The final volume is one-fourth of the initial volume. That means the volume decreases by

   $\frac{V_1 - V_2}{V_1} \times 100\% = \left(1 - \frac{1}{4}\right) \times 100\% = 75\%.$

   Watch out

   A common mistake is to use the relation $PV^\gamma = \text{constant}$ and then try to connect temperature via the ideal gas law, but that leads to the same result if done correctly. Another pitfall: forgetting that $\gamma - 1 = 0.5$ and accidentally using $\gamma$ directly.

   Tip

   When $\gamma = 1.5$, the exponent $\gamma-1 = 1/2$, so the relation $T \propto V^{-1/2}$ is easy to invert: doubling $T$ means $V$ must become $1/4$ of its original value.

   ---

   ✓Final answer

   The correct option is (D).

   ANSWER: D

6. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
   Q.A Carnot engine uses diatomic gas as a working substance. During the adiabatic expansion part of the cycle, if the volume of the gas becomes 32 times its initial volume, then the efficiency of the engine is (A) 100% (B) 75% (C) 50% (D) 25%

   ›Reveal solutionSolution

   The efficiency of a Carnot engine depends only on the temperatures of the reservoirs. For a diatomic gas undergoing adiabatic expansion with a volume ratio of 32, the temperature ratio is found using the adiabatic relation $TV^{\gamma-1} = \text{constant}$, with $\gamma = 7/5$. This gives an efficiency of 75%, so the correct option is (B).

   The key idea is that in a Carnot cycle, efficiency is $\eta = 1 - \frac{T_{\text{cold}}}{T_{\text{hot}}}$. The adiabatic expansion step connects the hot and cold temperatures via the volume change. For a diatomic gas, the adiabatic index $\gamma = C_p/C_v = 7/5$. Using the relation $T V^{\gamma-1} = \text{constant}$, we can find the temperature ratio from the given volume ratio, and then compute the efficiency.

   1. Recall the Carnot efficiency formula

      The efficiency of a Carnot engine is $\eta = 1 - \frac{T_c}{T_h}$, where $T_h$ is the temperature of the hot reservoir and $T_c$ is the temperature of the cold reservoir. The adiabatic expansion in the cycle connects these two temperatures.

   2. Apply the adiabatic relation for an ideal gas

      For a reversible adiabatic process, $T V^{\gamma-1} = \text{constant}$. Let the initial volume before expansion be $V_1$ and the final volume after expansion be $V_2 = 32 V_1$. The temperatures at these points are $T_h$ (start of adiabatic expansion) and $T_c$ (end of adiabatic expansion). Thus:

   $T_h V_1^{\gamma-1} = T_c V_2^{\gamma-1}$

   Rearranging:

   $\frac{T_c}{T_h} = \left( \frac{V_1}{V_2} \right)^{\gamma-1}$

   3. Substitute the volume ratio and $\gamma$ for a diatomic gas For a diatomic gas, $\gamma = \frac{7}{5}$, so $\gamma - 1 = \frac{2}{5}$. The volume ratio $V_2/V_1 = 32$, so $V_1/V_2 = 1/32$. Therefore:

   $\frac{T_c}{T_h} = \left( \frac{1}{32} \right)^{2/5}$

   Now, $32 = 2^5$, so $32^{2/5} = (2^5)^{2/5} = 2^2 = 4$. Hence:

   $\frac{T_c}{T_h} = \frac{1}{4}$

   4. Compute the efficiency

   $\eta = 1 - \frac{T_c}{T_h} = 1 - \frac{1}{4} = \frac{3}{4} = 75\%$

   Tip

   Notice that $32 = 2^5$ and the exponent $2/5$ cancels neatly, giving a simple ratio of 1/4. This is a common trick in such problems — always express the volume ratio as a power of 2 when the exponent is a fraction.

   Watch out

   A common mistake is to use $\gamma = 5/3$ (for monatomic gas) instead of $7/5$ (for diatomic). That would give $(1/32)^{2/3} = 1/8$, leading to an efficiency of 87.5%, which is not among the options. Always check the gas type.

   ✓Final answer

   The correct option is (B).

   ANSWER: B

7. TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
   Q.A Carnot engine uses diatomic gas as a working substance. During the adiabatic expansion part of the cycle, if the volume of the gas becomes 32 times its initial volume, then the efficiency of the engine is (A) 100% (B) 50% (C) 75% (D) 25%

   ›Reveal solutionSolution

   For a Carnot engine with a diatomic gas, the adiabatic relation $TV^{\gamma-1} = \text{constant}$ links the temperature drop to the volume expansion. Using $\gamma = 7/5$ for a diatomic gas and a 32‑fold volume increase in the adiabatic expansion, the efficiency comes out to 75%.

   The key idea is that a Carnot engine’s efficiency depends only on the temperatures of the two reservoirs: $\eta = 1 - T_C/T_H$. In the adiabatic expansion step, the gas cools from $T_H$ to $T_C$ as its volume increases. For an ideal gas undergoing a reversible adiabatic process, temperature and volume are related by $TV^{\gamma-1} = \text{constant}$. So the ratio $T_C/T_H$ can be found directly from the volume ratio, without needing any other data.

   A diatomic gas at ordinary temperatures has $\gamma = C_p/C_v = 7/5 = 1.4$. That’s a standard result: $C_v = \frac{5}{2}R$, $C_p = \frac{7}{2}R$, so $\gamma-1 = 2/5 = 0.4$.

   Now let’s work through the calculation.

   1. Adiabatic relation for the expansion For a reversible adiabatic process, $T V^{\gamma-1} = \text{constant}$. If the gas starts at temperature $T_H$ and volume $V_1$, and ends at $T_C$ and volume $V_2$, then

   $T_H V_1^{\gamma-1} = T_C V_2^{\gamma-1}.$

   Rearranging gives

   $\frac{T_C}{T_H} = \left(\frac{V_1}{V_2}\right)^{\gamma-1}.$

   2. Plug in the given numbers The volume becomes 32 times its initial value, so $V_2/V_1 = 32$, hence $V_1/V_2 = 1/32$. For a diatomic gas, $\gamma-1 = 2/5 = 0.4$. Therefore

   $\frac{T_C}{T_H} = \left(\frac{1}{32}\right)^{0.4}.$

   3. Simplify the exponent Write $32 = 2^5$. Then

   $\left(\frac{1}{2^5}\right)^{0.4} = 2^{-5 \times 0.4} = 2^{-2} = \frac{1}{4}.$

   So $T_C/T_H = 1/4$.

   4. Efficiency of the Carnot engine

   $\eta = 1 - \frac{T_C}{T_H} = 1 - \frac{1}{4} = \frac{3}{4} = 75\%.$

   Watch out

   A common mistake is to use $\gamma = 5/3$ (monatomic gas) instead of $7/5$ (diatomic). That would give $T_C/T_H = (1/32)^{2/3} = 1/8$, leading to $\eta = 87.5\%$, which is not among the options. Always check the gas’s atomicity.

   Tip

   Notice that $32 = 2^5$ and $\gamma-1 = 2/5$ for a diatomic gas. The exponents cancel neatly: $5 \times 2/5 = 2$. This is a common exam trick — the numbers are chosen to give a clean result. If you spot it, you can do the whole calculation mentally.

   ✓Final answer

   The efficiency of the engine is 75%, which corresponds to option (C).

8. TG EAPCET 2025Set ap-2025-04-29-FN1 markMCQ
   Q.The energy required to increase the radius of a soap bubble from 3 cm to 4 cm is (Surface tension of soap solution $= 3 \times 10^{-2}~\mathrm{Nm}^{-1}$) (A) $528 \times 10^{-6}~\mathrm{J}$ (B) $264 \times 10^{-6}~\mathrm{J}$ (C) $1056 \times 10^{-6}~\mathrm{J}$ (D) $478 \times 10^{-6}~\mathrm{J}$

   ›Reveal solutionSolution

   The energy required to expand a soap bubble is the work done against surface tension to increase its total surface area (considering both inner and outer surfaces). This energy is calculated as the product of surface tension and the change in total surface area. For the given radii, the energy required is $\boxed{528 \times 10^{-6}~\mathrm{J}}$.

   Concept and Intuition

   When a liquid surface expands, molecules from the bulk of the liquid move to the surface. To do this, they must overcome the cohesive forces pulling them inwards, which requires energy. This energy is stored as potential energy in the expanded surface, and the work done to create this new surface area is directly proportional to the increase in area. This phenomenon is quantified by surface tension ($T$), which can be defined as the work done per unit increase in surface area.

   For a soap bubble, a critical detail is that it has two free surfaces in contact with air: an inner surface and an outer surface. Therefore, when the radius of the bubble increases, both these surfaces expand. The total surface area that changes is twice the geometric surface area of a sphere.

   The energy required to increase the surface area of a liquid film by $\Delta A$ is given by:

   $W = T \times \Delta A$

   where $W$ is the work done (energy required), $T$ is the surface tension, and $\Delta A$ is the change in total surface area.

   Step-by-step Derivation

   1. Identify the given values and convert units:

      • Surface tension of soap solution, $T = 3 \times 10^{-2}~\mathrm{Nm}^{-1}$.
      • Initial radius of the soap bubble, $R_1 = 3~\mathrm{cm} = 3 \times 10^{-2}~\mathrm{m}$.
      • Final radius of the soap bubble, $R_2 = 4~\mathrm{cm} = 4 \times 10^{-2}~\mathrm{m}$.

   2. Calculate the initial total surface area ($A_1$) of the soap bubble:

      A spherical soap bubble has two surfaces (inner and outer). The surface area of a single sphere is $4\pi R^2$. Therefore, the total surface area of a soap bubble is $2 \times 4\pi R^2 = 8\pi R^2$.

      $A_1 = 8\pi R_1^2 = 8\pi (3 \times 10^{-2}~\mathrm{m})^2$

      $A_1 = 8\pi (9 \times 10^{-4}~\mathrm{m}^2)$

      $A_1 = 72\pi \times 10^{-4}~\mathrm{m}^2$

   3. Calculate the final total surface area ($A_2$) of the soap bubble:

      Similarly, for the final radius:

      $A_2 = 8\pi R_2^2 = 8\pi (4 \times 10^{-2}~\mathrm{m})^2$

      $A_2 = 8\pi (16 \times 10^{-4}~\mathrm{m}^2)$

      $A_2 = 128\pi \times 10^{-4}~\mathrm{m}^2$

   4. Calculate the change in total surface area ($\Delta A$):

      $\Delta A = A_2 - A_1$

      $\Delta A = (128\pi \times 10^{-4}~\mathrm{m}^2) - (72\pi \times 10^{-4}~\mathrm{m}^2)$

      $\Delta A = (128 - 72)\pi \times 10^{-4}~\mathrm{m}^2$

      $\Delta A = 56\pi \times 10^{-4}~\mathrm{m}^2$

   5. Calculate the energy required (work done, $W$):

      Using the formula $W = T \times \Delta A$:

      $W = (3 \times 10^{-2}~\mathrm{Nm}^{-1}) \times (56\pi \times 10^{-4}~\mathrm{m}^2)$

      $W = 3 \times 56\pi \times 10^{-2} \times 10^{-4}~\mathrm{J}$

      $W = 168\pi \times 10^{-6}~\mathrm{J}$

   6. Substitute the value of $\pi$ and calculate the numerical result:

      Using $\pi \approx 3.14159$:

      $W = 168 \times 3.14159 \times 10^{-6}~\mathrm{J}$

      $W \approx 527.787 \times 10^{-6}~\mathrm{J}$

      Rounding to the nearest whole number as per the options:

      $W \approx 528 \times 10^{-6}~\mathrm{J}$

   Watch out

   A common mistake is to consider only one surface of the soap bubble, using $4\pi R^2$ instead of $8\pi R^2$. This would lead to half the correct energy value. Always remember a soap bubble has two free surfaces.

   The calculated energy matches option (A).

   ✓Final answer

   The energy required to increase the radius of the soap bubble from 3 cm to 4 cm is $\boxed{528 \times 10^{-6}~\mathrm{J}}$.

9. TG EAPCET 2024Set ap-2024-05-08-FN1 markMCQ
   Q.The initial and the final temperatures of a black body are $27\,^\circ \mathrm{C}$ and $177\,^\circ \mathrm{C}$ respectively. The increase in the amount of radiation emitted per second is (A) 506.25% (B) 150.25% (C) 225.75% (D) 406.25%

   ›Reveal solutionSolution

   The radiation emitted by a black body scales as $T^4$ (Stefan–Boltzmann law). Converting temperatures to Kelvin and taking the ratio gives a factor of $(450/300)^4 = (1.5)^4 = 5.0625$, so the increase is $406.25\%$ — option (D).

   The key idea is Stefan’s law: for a black body, the total power radiated per unit area is $\sigma T^4$, where $T$ is the absolute temperature in Kelvin. When the temperature changes, the radiated power changes as the fourth power of the ratio of absolute temperatures. A common mistake is to use Celsius directly — that breaks the $T^4$ law because it’s not a ratio scale. Always convert to Kelvin first.

   1. Convert temperatures to Kelvin.

      $T_1 = 27^\circ\mathrm{C} + 273 = 300\ \mathrm{K}$

      $T_2 = 177^\circ\mathrm{C} + 273 = 450\ \mathrm{K}$

   2. Apply Stefan–Boltzmann law.

      The power radiated per second (for the same surface area) is proportional to $T^4$. So

   $\frac{P_2}{P_1} = \left(\frac{T_2}{T_1}\right)^4 = \left(\frac{450}{300}\right)^4 = (1.5)^4$

   Compute: $1.5^2 = 2.25$, then $2.25^2 = 5.0625$. Hence $P_2 = 5.0625\,P_1$.

   3. Find the percentage increase. Increase in radiation = $P_2 - P_1 = (5.0625 - 1)P_1 = 4.0625\,P_1$. As a percentage of the original: $4.0625 \times 100\% = 406.25\%$.

   Watch out

   Do not use Celsius temperatures in the ratio. $(177/27)^4$ gives a huge wrong number — the $T^4$ law only holds for absolute temperature.

   Tip

   $(1.5)^4 = (3/2)^4 = 81/16 = 5.0625$ is a clean fraction — useful for quick mental calculation.

   ✓Final answer

   The increase in radiation emitted per second is $406.25\%$, which corresponds to option (D).

10. TG EAPCET 2024Set ap-2024-05-07-FN1 markMCQ
    Q.A black body at a temperature of $125^\circ \mathrm{C}$, emits heat at the rate of $32~\mathrm{Wm}^{-2}$. The rate of heat emitted by the body when the temperature of the body is increased by $398~\mathrm{K}$ is (A) $64~\mathrm{Wm}^{-2}$ (B) $128~\mathrm{Wm}^{-2}$ (C) $512~\mathrm{Wm}^{-2}$ (D) $256~\mathrm{Wm}^{-2}$

    ›Reveal solutionSolution

    The key idea is that the power radiated by a black body is proportional to the fourth power of its absolute temperature (Stefan–Boltzmann law). Converting the given Celsius temperature to Kelvin and applying the ratio of powers gives the new emission rate as $512~\mathrm{Wm}^{-2}$, so the correct option is (C).

    Concept and intuition

    A black body is an ideal emitter: the power it radiates per unit area depends only on its absolute temperature $T$ (in Kelvin). The Stefan–Boltzmann law states:

    $P = \sigma T^4$

    where $\sigma$ is the Stefan–Boltzmann constant.

    If the temperature changes, the new power is proportional to the fourth power of the new absolute temperature. The problem gives an initial temperature in Celsius and a temperature increase in Kelvin (which is the same as an increase in Celsius degrees). The crucial step is to work entirely in Kelvin.

    Step-by-step solution

    1. Convert the initial temperature to Kelvin The initial temperature is $125^\circ\mathrm{C}$.

    $T_1 = 125 + 273 = 398~\mathrm{K}$

    2. Identify the temperature increase The temperature is increased by $398~\mathrm{K}$. So the new absolute temperature is:

    $T_2 = T_1 + 398 = 398 + 398 = 796~\mathrm{K}$

    3. Apply the Stefan–Boltzmann law The initial power per unit area is $P_1 = 32~\mathrm{Wm}^{-2}$. Since $P \propto T^4$, we have:

    $\frac{P_2}{P_1} = \left( \frac{T_2}{T_1} \right)^4$

    4. Compute the ratio

    $\frac{T_2}{T_1} = \frac{796}{398} = 2$

    Therefore:

    $\frac{P_2}{P_1} = 2^4 = 16$

    5. Find the new power

    $P_2 = 16 \times P_1 = 16 \times 32 = 512~\mathrm{Wm}^{-2}$

    Watch out

    A common mistake is to forget that the Stefan–Boltzmann law requires absolute temperature. Using Celsius directly would give a wrong ratio. Also, note that an increase of $398~\mathrm{K}$ is exactly the same as an increase of $398^\circ\mathrm{C}$, but the starting temperature must be in Kelvin.

    Tip

    Notice that the initial temperature in Kelvin (398 K) equals the temperature increase (398 K). This makes the new temperature exactly double the old one, so the power multiplies by $2^4 = 16$. This neat coincidence simplifies the arithmetic.

    ✓Final answer

    The correct option is (C).

    ANSWER: C

11. TG EAPCET 2024Set eng-2024-05-10-AN1 markMCQ
    Q.Steam of mass 60 g at a temperature 100 $^\circ$C is mixed with water of mass 360 g at a temperature 40 $^\circ$C. The ratio of the masses of steam and water in equilibrium is (Latent heat of steam is 540 cal g$^{-1}$ and specific heat capacity of water is 1 cal g$^{-1}$ $^\circ$C$^{-1}$) (A) 1 : 20 (B) 1 : 10 (C) 1 : 5 (D) 1 : 3

    ›Reveal solutionSolution

    Warming the water to $100^\circ\text{C}$ condenses only $40\text{ g}$ of steam, leaving $20\text{ g}$ steam and $400\text{ g}$ water — a ratio of $1:20$.

    Heat needed to raise the water to $100^\circ\text{C}$.

    The $360\text{ g}$ of water at $40^\circ\text{C}$ needs to be heated by $60^\circ\text{C}$:

    $Q_{\text{water}} = m\,c\,\Delta T = 360 \times 1 \times (100-40) = 21600 \text{ cal}.$

    Heat available from condensing all the steam.

    Condensing the entire $60\text{ g}$ of steam would release

    $Q_{\text{steam}} = 60 \times 540 = 32400 \text{ cal}.$

    Since $32400 > 21600$, not all the steam condenses. Equilibrium is reached at $100^\circ\text{C}$ with steam and water coexisting.

    Mass of steam that condenses.

    Let $m$ grams of steam condense, supplying exactly the heat the water needs:

    $m \times 540 = 21600 \quad\Rightarrow\quad m = 40 \text{ g}.$

    Masses at equilibrium.

    • Steam remaining $= 60 - 40 = 20 \text{ g}$.
    • Water $= 360 + 40 = 400 \text{ g}$ (original water plus condensed steam).

    $\text{steam} : \text{water} = 20 : 400 = 1 : 20.$

    ✓Final answer

    The ratio of steam to water at equilibrium is $1:20$ — option (A).

12. TG EAPCET 2024Set eng-2024-05-10-AN1 markMCQ
    Q.The temperature at which the rms speed of oxygen molecules is 75% of rms speed of nitrogen molecules at a temperature of 287 $^\circ$C is (A) 87 $^\circ$C (B) 127 $^\circ$C (C) 227 $^\circ$C (D) 360 $^\circ$C

    ›Reveal solutionSolution

    The rms speed of a gas depends on temperature and molar mass; setting the oxygen rms speed to 75% of nitrogen’s rms speed gives a relation that yields the required temperature. The answer is 87 °C.

    The key idea is that the root‑mean‑square speed of gas molecules is given by

    $v_{\text{rms}} = \sqrt{\frac{3RT}{M}}$

    where $T$ is the absolute temperature (in kelvin) and $M$ is the molar mass (in kg mol⁻¹).

    We are told that for oxygen, $v_{\text{rms,O}_2}$ is 75% of $v_{\text{rms,N}_2}$ at a known temperature of nitrogen. This lets us set up a ratio that eliminates the gas constant $R$ and directly relates the temperatures and molar masses.

    1. Convert the given nitrogen temperature to kelvin The nitrogen temperature is $287\,^\circ\text{C}$.

    $T_{\text{N}_2} = 287 + 273 = 560\ \text{K}$

    2. Write the rms speed expressions For nitrogen (molar mass $M_{\text{N}_2} = 28\ \text{g mol}^{-1} = 0.028\ \text{kg mol}^{-1}$):

    $v_{\text{rms,N}_2} = \sqrt{\frac{3R \cdot 560}{0.028}}$

    For oxygen (molar mass $M_{\text{O}_2} = 32\ \text{g mol}^{-1} = 0.032\ \text{kg mol}^{-1}$) at unknown temperature $T_{\text{O}_2}$:

    $v_{\text{rms,O}_2} = \sqrt{\frac{3R \cdot T_{\text{O}_2}}{0.032}}$

    3. Apply the given condition

    $v_{\text{rms,O}_2} = 0.75 \cdot v_{\text{rms,N}_2}$

    Substitute the expressions:

    $\sqrt{\frac{3R\,T_{\text{O}_2}}{0.032}} = 0.75 \cdot \sqrt{\frac{3R \cdot 560}{0.028}}$

    4. Cancel common factors and square both sides The factor $3R$ cancels from both sides:

    $\sqrt{\frac{T_{\text{O}_2}}{0.032}} = 0.75 \cdot \sqrt{\frac{560}{0.028}}$

    Squaring:

    $\frac{T_{\text{O}_2}}{0.032} = (0.75)^2 \cdot \frac{560}{0.028}$

    $\frac{T_{\text{O}_2}}{0.032} = 0.5625 \cdot \frac{560}{0.028}$

    5. Simplify the right‑hand side First compute $\frac{560}{0.028}$:

    $\frac{560}{0.028} = \frac{560}{28/1000} = 560 \times \frac{1000}{28} = 20 \times 1000 = 20000$

    Then multiply by 0.5625:

    $0.5625 \times 20000 = 11250$

    So we have:

    $\frac{T_{\text{O}_2}}{0.032} = 11250$

    6. Solve for $T_{\text{O}_2}$

    $T_{\text{O}_2} = 11250 \times 0.032 = 360\ \text{K}$

    7. Convert back to Celsius

    $T_{\text{O}_2} (^\circ\text{C}) = 360 - 273 = 87\,^\circ\text{C}$

    Watch out

    A common mistake is to forget to convert Celsius to kelvin before using the rms formula, or to use molar masses in grams per mole without converting to kilograms. Both errors lead to wrong numerical factors.

    Tip

    Notice that the ratio of rms speeds depends only on the square root of the ratio of temperatures divided by molar masses. This shortcut lets you skip writing the full expressions each time.

    ✓Final answer

    The correct option is (A).

    ANSWER: A