Ice at -5^{\circ} {\C} is heated to become vapour with temperature of 110^{\circ} {\C} at atmospheric pressure. The entropy change associated with this process can be obtained from [23 Jan 2025 Shift 1]

Correct Answer is : 273K∫268kCp⋅mTdT+∆Hm, fusion Tf+∆Hm, vaporisation Tb+373K∫273KCp,mTdT+383K∫373KCp,mT

\int_{269 K}^{273 K} C_{p, m} d T+\frac{∆ H_m, \text{ fusion }}{T_f}+\frac{∆ H_m, \text{ vaporisation }}{T_b}+\int_{273 K}^{373 K} C_{p, m} d T+\int_{3 \/ 3 K}^{333 K} C_{p, m} d T

\int_{268 {\K}}^{383 {\K}} {\C}_{ {\p}} {\dT}+ {∆ {\H}_{\text{melting }}}/{273}+ {∆ {\H}_{\text{boiling }}}/{373}

\int_{268 k}^{273 K} \frac{C_p \cdot m}{T} d T+\frac{∆ H_m, \text{ fusion }}{T_f}+\frac{∆ H_m, \text{ vaporisation }}{T_b}+\int_{273 K}^{373 K} \frac{C_p, m}{T} d T+\int_{373 K}^{383 K} \frac{C_p, m}{T}

\int_{268 {\K}}^{383 {\K}} {\C}_{ {\p}} {\dT}+ { {\q}_{ {\rev}}}/{ {\T}}

JEE Main 23 Jan 2025 Shift 1 Paper

Section: Chemistry

Question : 54 of 75

Marks: +1, -0

Ice at −5∘C is heated to become vapour with temperature of 110∘C at atmospheric pressure. The entropy change associated with this process can be obtained from

[23 Jan 2025 Shift 1]

273K
∫
269K
Cp,mdT+‌
∆Hm,‌ fusion ‌
Tf
+‌
∆Hm,‌ vaporisation ‌
Tb
+
373K
∫
273K
Cp,mdT+
333K
∫
3∕3K
Cp,mdT
383K
∫
268K
Cp‌dT+‌
∆H‌melting ‌
273
+‌
∆H‌boiling ‌
373
273K
∫
268k
‌
Cp⋅m
T
dT+‌
∆Hm,‌ fusion ‌
Tf
+‌
∆Hm,‌ vaporisation ‌
Tb
+
373K
∫
273K
‌
Cp,m
T
dT+
383K
∫
373K
‌
Cp,m
T
383K
∫
268K
Cp‌dT+‌
qrev
T

Validate

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