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Thermodynamics question

2025 · 7 Apr · Shift 1 · Q3
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Thermodynamics question

2025 · 7 Apr · Shift 1 · Q3

JEE MainChemistryThermodynamicsMCQ+4 / −1
Total enthalpy change for freezing of 1 mol of water at 10∘C10^{\circ} \mathrm{C}10∘C to ice at −10∘C-10^{\circ} \mathrm{C}−10∘C is ‾\underline{\hspace{2cm}}​ (Given: Δfus H=x kJ/mol\Delta_{\text {fus }} \mathrm{H}=x \mathrm{~kJ} / \mathrm{mol}Δfus ​H=x kJ/mol Cp[H2O(l)]=y J mol−1 K−1Cp[H2O( s)]=z J mol−1 K−1\begin{aligned} & \mathrm{C}_{\mathrm{p}}\left[\mathrm{H}_2 \mathrm{O}(\mathrm{l})\right]=y \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} \\ & \mathrm{C}_{\mathrm{p}}\left[\mathrm{H}_2 \mathrm{O}(\mathrm{~s})\right]=z \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} \end{aligned}​Cp​[H2​O(l)]=y J mol−1 K−1Cp​[H2​O( s)]=z J mol−1 K−1​
  1. A
    −x−10y−10z-x-10 y-10 z−x−10y−10z
  2. B
    x−10y−10zx-10 y-10 zx−10y−10z
  3. C
    −10(100x+y+z)-10(100 x+y+z)−10(100x+y+z)
  4. D
    10(100x+y+z)10(100 \mathrm{x}+y+z)10(100x+y+z)
View written solutionFree

Correct answer: C

  1. Break the process into three steps for converting \mathrm{H_2O(l,\ 10^ C)} \to \mathrm{H_2O(s,\ -10^ C)}

    • Step 1: Cool liquid water from 10^ C to 0^ C
    • Step 2: Freeze water at 0^ C
    • Step 3: Cool ice from 0^ C to -10^ C

  2. Enthalpy change for step 1: cooling liquid water

    ΔH1=nCp(l)ΔT\Delta H_1 = n C_p(\text{l}) \Delta TΔH1​=nCp​(l)ΔT

    Here, n=1n=1n=1, Cp(l)=y J mol−1K−1C_p(\text{l})=y\ \mathrm{J\ mol^{-1}K^{-1}}Cp​(l)=y J mol−1K−1, and ΔT=0−10=−10 K\Delta T = 0-10 = -10\ \mathrm{K}ΔT=0−10=−10 K

    So, ΔH1=y(−10)=−10y J\Delta H_1 = y(-10) = -10y\ \mathrm{J}ΔH1​=y(−10)=−10y J

  3. Enthalpy change for step 2: freezing at 0^ C

    Given heat of fusion of water is ΔfusH=x kJ/mol\Delta_{\text{fus}}H = x\ \mathrm{kJ/mol}Δfus​H=x kJ/mol

    Fusion means melting, which is endothermic. Therefore, freezing is the reverse process: ΔH2=−x kJ=−1000x J\Delta H_2 = -x\ \mathrm{kJ} = -1000x\ \mathrm{J}ΔH2​=−x kJ=−1000x J

  4. Enthalpy change for step 3: cooling ice

    ΔH3=nCp(s)ΔT\Delta H_3 = n C_p(\text{s}) \Delta TΔH3​=nCp​(s)ΔT

    Here, Cp(s)=z J mol−1K−1C_p(\text{s})=z\ \mathrm{J\ mol^{-1}K^{-1}}Cp​(s)=z J mol−1K−1 and ΔT=−10−0=−10 K\Delta T = -10-0 = -10\ \mathrm{K}ΔT=−10−0=−10 K

    Hence, ΔH3=z(−10)=−10z J\Delta H_3 = z(-10) = -10z\ \mathrm{J}ΔH3​=z(−10)=−10z J

  5. Add all three contributions

    ΔH=ΔH1+ΔH2+ΔH3\Delta H = \Delta H_1 + \Delta H_2 + \Delta H_3ΔH=ΔH1​+ΔH2​+ΔH3​

    ΔH=−10y−1000x−10z\Delta H = -10y -1000x -10zΔH=−10y−1000x−10z

    Factor out −10-10−10: ΔH=−10(100x+y+z)\Delta H = -10(100x+y+z)ΔH=−10(100x+y+z)

  6. Match with the options

    This corresponds to: C: −10(100x+y+z)\boxed{\text{C: } -10(100x+y+z)}C: −10(100x+y+z)​

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