A vessel at − 5 ∘ C contains 100 gm ice at − 5 ∘ C . The vessel has water equivalent of 40 gm . List-I gives quantity and temperature of substance added. List-II gives final temperature and compositio

Question

A vessel at − 5 ∘ C contains 100 gm ice at − 5 ∘ C . The vessel has water equivalent of 40 gm . List-I gives quantity and temperature of substance added. List-II gives final temperature and composition of mixture. Match the column. List-I List-II P. 10 gm water at 3 0 ∘ C 1. Temperature > 0 ∘ C Q. 4 gm water at 1 0 ∘ C 2. Temperature = 0 ∘ C , Ice partly melts R. 1 gm steam at 10 0 ∘ C 3. Temperature < 0 ∘ C S. 20 gm steam at 10 0 ∘ C 4. Added water freezes partly. Temperature = 0 ∘ C Code:

TestHub · Accepted Answer

Q 0 ​ = 40 × 5 + 100 × 2 1 ​ × 5 = 450 cal Q a ll ​ = 450 + 8000 = 8450 cal P: partial freezing. 10 × 30 = 300 cal ( P → 4 ) Q: below freezing. 4 × 10 + 4 × 80 = 360 < 450 ( Q → 3 ) R: partial melting. 540 + 100 = 640 ( R → 2 ) S: above freezing. 20 ( 540 + 100 ) = 12800 > 8450 ( S → 1 )

List-I	List-II
P. $10 gm$ water at $3 0^{\circ} C$	1. Temperature $> 0^{\circ} C$
Q. $4 gm$ water at $1 0^{\circ} C$	2. Temperature $= 0^{\circ} C$ , Ice partly melts
R. $1 gm$ steam at $10 0^{\circ} C$	3. Temperature $< 0^{\circ} C$
S. $20 gm$ steam at $10 0^{\circ} C$	4. Added water freezes partly. Temperature $= 0^{\circ} C$

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JEE Advanced 2028 (Class 11)