Which of the following mixing of 1M base and 1M acid leads to the largest increase in temperature?
A.\text{30 mL HCl and 30 mL NaOH}
B.\text{30 mL } \mathrm{CH_{3}COOH} \text{ and 30 mL NaOH}
C.\text{50 mL HCl and 20 mL NaOH}
D.\text{45 mL } \mathrm{CH_{3}COOH} \text{ and 25 mL NaOH}
Solution & Explanation
### Related Formula
Q = n_textreacted cdot Delta H_textneutralization$Q = n_{\text{reacted}} \cdot \Delta H_{\text{neutralization}}$Delta T = fracQm cdot c$\Delta T = \frac{Q}{m \cdot c}$
### Core Logic
The temperature rise depends directly on the total heat released (Q$Q$) normalized by the total heat capacity of the resulting mixed volume (m cdot c$m \cdot c$). Let's evaluate the millimoles of mathrmH^+$\mathrm{H}^+$ and mathrmOH^-$\mathrm{OH}^-$ that react in each mixture:
1. **Option 1:** 30text mL of 1mathrmM mathrmHCl + 30text mL of 1mathrmM mathrmNaOH$30\text{ mL of } 1\mathrm{M } \mathrm{HCl} + 30\text{ mL of } 1\mathrm{M } \mathrm{NaOH}$textReactive millimoles = 30text mmol$\text{Reactive millimoles} = 30\text{ mmol}$. Both are strong electrolytes, releasing full neutralization energy (sim -57.3text kJ/mol$\sim -57.3\text{ kJ/mol}$).
Total volume = 60text mL$60\text{ mL}$.
2. **Option 2:** 30text mL of 1mathrmM mathrmCH_3COOH + 30text mL of 1mathrmM mathrmNaOH$30\text{ mL of } 1\mathrm{M } \mathrm{CH_3COOH} + 30\text{ mL of } 1\mathrm{M } \mathrm{NaOH}$textReactive millimoles = 30text mmol$\text{Reactive millimoles} = 30\text{ mmol}$. However, since acetic acid is a weak acid, part of the heat is consumed in its ionization. Thus, less total heat is evolved compared to Option 1.
3. **Option 3:** 50text mL of 1mathrmM mathrmHCl + 20text mL of 1mathrmM mathrmNaOH$50\text{ mL of } 1\mathrm{M } \mathrm{HCl} + 20\text{ mL of } 1\mathrm{M } \mathrm{NaOH}$textLimiting reagent = mathrmNaOH = 20text mmol$\text{Limiting reagent} = \mathrm{NaOH} = 20\text{ mmol}$. Only 20text mmol$20\text{ mmol}$ reacts.
Total volume = 70text mL$70\text{ mL}$.
4. **Option 4:** 45text mL of 1mathrmM mathrmCH_3COOH + 25text mL of 1mathrmM mathrmNaOH$45\text{ mL of } 1\mathrm{M } \mathrm{CH_3COOH} + 25\text{ mL of } 1\mathrm{M } \mathrm{NaOH}$textLimiting reagent = 25text mmol$\text{Limiting reagent} = 25\text{ mmol}$ weak neutralization profile.
Comparing Option 1 and Option 3, Option 1 releases significantly more heat (30text mmol$30\text{ mmol}$ vs 20text mmol$20\text{ mmol}$) into a smaller volume (60text mL$60\text{ mL}$ vs 70text mL$70\text{ mL}$), yielding the largest increase in temperature Delta T$\Delta T$.
### Pattern Recognition
To maximize Delta T$\Delta T$, look for the option that maximizes the amount of reacting strong acid and strong base equivalents while keeping the total solution volume as small as possible.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Thermodynamics
Class 11 Chemistry: Equilibrium
Keywords:#heat of neutralization#maximum temperature rise acid base#JEE Main 2025 Evening Q31#Thermodynamics calorimetry
More Thermodynamics Previous-Year Questions — Page 3
Q62025Heat and Work in Thermodynamic Processes
A poly-atomic molecule (C_V = 3R, C_P = 4R$C_V = 3R, C_P = 4R$, where R$R$ is gas constant) goes from phase space point A(P_A = 10^5mathrm~Pa, V_A = 4 times 10^-6mathrm~m^3)$A(P_A = 10^5\mathrm{~Pa}, V_A = 4 \times 10^{-6}\mathrm{~m}^3)$ to point B(P_B = 5 times 10^4mathrm~Pa, V_B = 6 times 10^-6mathrm~m^3)$B(P_B = 5 \times 10^4\mathrm{~Pa}, V_B = 6 \times 10^{-6}\mathrm{~m}^3)$ to point C(P_C = 10^4mathrm~Pa, V_C = 8 times 10^-6mathrm~m^3)$C(P_C = 10^4\mathrm{~Pa}, V_C = 8 \times 10^{-6}\mathrm{~m}^3)$. A$A$ to B$B$ is an adiabatic path and B$B$ to C$C$ is an isothermal path. The net heat absorbed per unit mole by the system is:
The graph depicts a pressure vs volume plot indicating paths from state A to B (adiabatic) and from B to C (isothermal).
### Related Formula
Delta Q_textnet = Delta Q_AB + Delta Q_BC$\Delta Q_{\text{net}} = \Delta Q_{AB} + \Delta Q_{BC}$Delta Q_textisothermal = nRT lnleft(fracV_fV_i
ight) = P_i V_i lnleft(fracV_fV_i
ight)$\Delta Q_{\text{isothermal}} = nRT \ln\left(\frac{V_f}{V_i}
ight) = P_i V_i \ln\left(\frac{V_f}{V_i}
ight)$
### Core Logic
For path A rightarrow B$A \rightarrow B$:
Since it is given as an adiabatic path:
Delta Q_AB = 0$\Delta Q_{AB} = 0$
For path B rightarrow C$B \rightarrow C$:
Since it is given as an isothermal path, the change in internal energy Delta U_BC = 0$\Delta U_{BC} = 0$. From the first law of thermodynamics, heat absorbed equals work done:
Delta Q_BC = W_BC = nRT_B lnleft(fracV_CV_B
ight)$\Delta Q_{BC} = W_{BC} = nRT_B \ln\left(\frac{V_C}{V_B}
ight)$
Using the ideal gas state at point B$B$, nRT_B = P_B V_B$nRT_B = P_B V_B$:
P_B V_B = (5 times 10^4 mathrm~Pa) times (6 times 10^-6 mathrm~m^3) = 0.3 mathrm~J$P_B V_B = (5 \times 10^4 \mathrm{~Pa}) \times (6 \times 10^{-6} \mathrm{~m}^3) = 0.3 \mathrm{~J}$
Wait, let's express it in terms of the gas constant R$R$ for a single mole (n=1$n=1$) using temperature data directly provided in the original figure labels (T_B = 450mathrm~K$T_B = 450\mathrm{~K}$):
Delta Q_BC = (1) cdot R cdot (450) cdot lnleft(frac8 times 10^-66 times 10^-6right)$\Delta Q_{BC} = (1) \cdot R \cdot (450) \cdot \ln\left(\frac{8 \times 10^{-6}}{6 \times 10^{-6}}\right)$Delta Q_BC = 450 R lnleft(frac43
ight) = 450 R (ln 4 - ln 3)$\Delta Q_{BC} = 450 R \ln\left(\frac{4}{3}
ight) = 450 R (\ln 4 - \ln 3)$
Thus, the total net heat absorbed per unit mole is:
Delta Q = 0 + 450 R (ln 4 - ln 3) = 450 R (ln 4 - ln 3)$\Delta Q = 0 + 450 R (\ln 4 - \ln 3) = 450 R (\ln 4 - \ln 3)$
### Pattern Recognition
Adiabatic paths have zero heat exchange by baseline definition. The calculation boils down directly to the work done during the isothermal stage B rightarrow C$B \rightarrow C$ matching RT ln(V_f/V_i)$RT \ln(V_f/V_i)$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Thermodynamics
### Core Logic
Let's treat the given parameters as terms for exothermic heats evolved on the product side:
1) C(textdiamond)
ightarrow C(textgraphite)$C(\text{diamond})
ightarrow C(\text{graphite})$, Delta H = -X$Delta H = -X$
2) C(textdiamond) + O_2(g)
ightarrow CO_2(g)$C(\text{diamond}) + O_{2}(g)
ightarrow CO_{2}(g)$, Delta H = -Y$Delta H = -Y$
3) C(textgraphite) + O_2(g)
ightarrow CO_2(g)$C(\text{graphite}) + O_{2}(g)
ightarrow CO_{2}(g)$, Delta H = -Z$Delta H = -Z$
By subtracting Equation (3) from Equation (2):
C(textdiamond) - C(textgraphite)
ightarrow 0 implies C(textdiamond)
ightarrow C(textgraphite)$C(\text{diamond}) - C(\text{graphite})
ightarrow 0 implies C(\text{diamond})
ightarrow C(\text{graphite})$Delta H = (-Y) - (-Z) = Z - Y$Delta H = (-Y) - (-Z) = Z - Y$
Matching this to Equation (1):
-X = Z - Y implies X = Y - Z$-X = Z - Y implies X = Y - Z$
### Pattern Recognition
Hess's Law states that the enthalpy change of an overall reaction is equal to the sum of the enthalpy changes of its individual steps.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Chemistry: Thermodynamics
Q72025Carnot Engine and Efficiency
A Carnot engine (E) is working between two temperatures 473K and 273K. In a new system two engines - engine mathrmE_1$\mathrm{E}_1$ works between 473K to 373K and engine mathrmE_2$\mathrm{E}_2$ works between 373K to 273K. If eta_12$\eta_{12}$ , eta_1$\eta_1$ and eta_2$\eta_2$ are the efficiencies of the engines E, mathrmE_1$\mathrm{E}_1$ and mathrmE_2$\mathrm{E}_2$ , respectively, then
### Related Formula
eta = 1 - fracmathrmT_LmathrmT_H$\eta = 1 - \frac{\mathrm{T}_L}{\mathrm{T}_H}$
### Core Logic
Let's compute the efficiency parameters explicitly:
eta_12 = 1 - frac273473 = frac200473 approx 0.423$\eta_{12} = 1 - \frac{273}{473} = \frac{200}{473} \approx 0.423$eta_1 = 1 - frac373473 = frac100473 approx 0.211$\eta_1 = 1 - \frac{373}{473} = \frac{100}{473} \approx 0.211$eta_2 = 1 - frac273373 = frac100373 approx 0.268$\eta_2 = 1 - \frac{273}{373} = \frac{100}{373} \approx 0.268$
Evaluating the linear sum of fractional bounds:
eta_1 + eta_2 = 0.211 + 0.268 = 0.479$\eta_1 + \eta_2 = 0.211 + 0.268 = 0.479$
Comparing the outputs clearly demonstrates:
eta_12 < eta_1 + eta_2$\eta_{12} < \eta_1 + \eta_2$
### Step 1: Final Conclusion
Thus, the inequality satisfies option (1).
### Pattern Recognition
The joint efficiency of cascading perfect thermodynamic steps is bounded multiplicatively as (1-eta_12) = (1-eta_1)(1-eta_2)$(1-\eta_{12}) = (1-\eta_1)(1-\eta_2)$, which algebraically forces eta_12 = eta_1 + eta_2 - eta_1eta_2 < eta_1 + eta_2$\eta_{12} = \eta_1 + \eta_2 - \eta_1\eta_2 < \eta_1 + \eta_2$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Thermodynamics
Q12025Phase Change and Melting
During the melting of a slab of ice at 273mathrm~K$273\mathrm{~K}$ at atmospheric pressure:
A. Internal energy of ice-water system remains unchanged.
B. Positive work is done by the ice-water system on the atmosphere.
C. Internal energy of the ice-water system decreases.
D. Positive work is done on the ice-water system by the atmosphere.
Solution
### Related Formula
Delta U = Delta Q + Delta W_texton system$\Delta U = \Delta Q + \Delta W_{\text{on system}}$
where,
Delta U$\Delta U$ = change in internal energy,
Delta Q$\Delta Q$ = heat exchange,
Delta W_texton system$\Delta W_{\text{on system}}$ = work done on the system.
### Core Logic
During the melting of ice at 273mathrm~K$273\mathrm{~K}$, the density of water is higher than the density of ice. This means the volume of the ice-water system decreases during melting:
V_f < V_i implies Delta V < 0$V_f < V_i \implies \Delta V < 0$
Since the system contracts, the atmosphere performs positive work on it:
W_texton system = -P Delta V > 0$W_{\text{on system}} = -P \Delta V > 0$
Additionally, heat is absorbed by the system to melt the ice, so Delta Q > 0$\Delta Q > 0$. By the first law of thermodynamics, since both Delta Q$\Delta Q$ and Delta W_texton system$\Delta W_{\text{on system}}$ are positive, the internal energy of the system increases:
Delta U = Delta Q + W_texton system > 0$\Delta U = \Delta Q + W_{\text{on system}} > 0$
### Step 1: Evaluation of Options
Let's check the given options:
1. Internal energy remains unchanged rightarrow$\rightarrow$ False (it increases).
2. Positive work is done by the system rightarrow$\rightarrow$ False (work done by the system is negative since it contracts).
3. Internal energy decreases rightarrow$\rightarrow$ False.
4. Positive work is done on the ice-water system by the atmosphere rightarrow$\rightarrow$ True (since volume decreases under atmospheric pressure).
### Pattern Recognition
Remember: Ice contracts upon melting (unlike most solids). Shrinking volume (V downarrow$V \downarrow$) under positive pressure (P > 0$P > 0$) means the surroundings (atmosphere) compress it, performing positive work on the system.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Thermal Properties of Matter
Class 11 Physics: Thermodynamics
Q112025Isothermal Expansion with Non-Linear Spring
A piston of mass M$M$ is hung from a massless spring whose restoring force law goes as F = -kx^3$F = -kx^3$, where k$k$ is the spring constant of appropriate dimension. The piston separates the vertical chamber into two parts, where the bottom part is filled with 'n$n$' moles of an ideal gas. An external work is done on the gas isothermally (at a constant temperature T$T$) with the help of a heating filament (with negligible volume) mounted in lower part of the chamber, so that the piston goes up from a height L_0$L_0$ to L_1$L_1$, the total energy delivered by the filament is (Assume spring to be in its natural length before heating)
A schematic of a piston of mass M connected to a spring inside a vertical chamber, separating gas at the bottom from vacuum/atmosphere at the top.
### Related Formula
First Law of Thermodynamics:
Delta Q = Delta U + W_textby gas$\Delta Q = \Delta U + W_{\text{by gas}}$
Work done by an ideal gas during isothermal expansion:
W_textgas = nRTlnleft(fracV_1V_0right) = nRTlnleft(fracL_1L_0
ight)$W_{\text{gas}} = nRT\ln\left(\frac{V_1}{V_0}\right) = nRT\ln\left(\frac{L_1}{L_0}
ight)$
Conservation of Energy (Work-Energy Theorem):
Total energy delivered by the heating filament (W_textfilament$W_{\text{filament}}$) must equal the total work needed to lift the piston against gravity and compress the non-linear spring.
### Core Logic
Since the process is isothermal, the change in internal energy of the ideal gas is zero (Delta U = 0$\Delta U = 0$). Hence:
Q = W_textgas$Q = W_{\text{gas}}$
By the Work-Energy Theorem for the piston:
W_textgas + W_textfilament = Delta U_textgravity + Delta U_textspring$W_{\text{gas}} + W_{\text{filament}} = \Delta U_{\text{gravity}} + \Delta U_{\text{spring}}$
Let's evaluate each term:
- Increase in gravitational potential energy:
Delta U_textgravity = Mg(L_1 - L_0)$\Delta U_{\text{gravity}} = Mg(L_1 - L_0)$
- Increase in spring potential energy:
U_textspring = -int_L_0^L_1 F_textrestoring dx = int_L_0^L_1 kx^3 dx = frack4(L_1^4 - L_0^4)$U_{\text{spring}} = -\int_{L_0}^{L_1} F_{\text{restoring}} dx = \int_{L_0}^{L_1} kx^3 dx = \frac{k}{4}(L_1^4 - L_0^4)$
### Step 1: Finding Total Energy Delivered
Isolating W_textfilament$W_{\text{filament}}$ (the net external energy delivered to the gas system):
W_textfilament = W_textgas + Mg(L_1 - L_0) + frack4(L_1^4 - L_0^4)$W_{\text{filament}} = W_{\text{gas}} + Mg(L_1 - L_0) + \frac{k}{4}(L_1^4 - L_0^4)$
Since W_textgas = nRTlnleft(fracL_1L_0
ight)$W_{\text{gas}} = nRT\ln\left(\frac{L_1}{L_0}
ight)$:
W_textfilament = nRTlnleft(fracL_1L_0
ight) + Mg(L_1 - L_0) + frack4(L_1^4 - L_0^4)$W_{\text{filament}} = nRT\ln\left(\frac{L_1}{L_0}
ight) + Mg(L_1 - L_0) + \frac{k}{4}(L_1^4 - L_0^4)$
### Pattern Recognition
Notice how energy conservation instantly frames this complex thermodynamics question. The heating filament's energy simply goes into three distinct stores: the isothermal work of gas expansion, raising the mass against gravity (Mgh$Mgh$), and the potential energy of the spring (integrated from kx^3$kx^3$). Keeping this total energy ledger in mind prevents tedious mathematical tangents.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Thermodynamics: First Law
Class 11 Physics: Work, Energy and Power: Variable Force Integration
More Thermodynamics Questions — jee_main_2025_24_jan_evening
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