Suppose we have ice at 0°C, water at 50°C and steam at 100°C. Some amount of each is taken and mixed as described in List-I. For each situation, find the final temperature of the mixture out of possib

Question

Suppose we have ice at 0°C, water at 50°C and steam at 100°C. Some amount of each is taken and mixed as described in List-I. For each situation, find the final temperature of the mixture out of possible values in List-II. Assume that there is no heat loss to surroundings. Latent heat of fusion of ice $= 80 cal / gm$ , latent heat of vapourisation of water $= 540 cal / gm$ , specific heat of water $= 1 cal / gm^{\circ} C$ .

List - I	List - II
(I) 20 gm ice + 40 gm water	(P) $0^{\circ} C$
(II) 50 gm ice + 10 gm steam	(Q) $6.6 7^{\circ} C$
(III) 100 gm water + 10 gm steam	(R) $4 0^{\circ} C$
(IV) 60 gm ice + 16 gm water + 5 gm steam	(S) $5 0^{\circ} C$
	(T) $10 0^{\circ} C$

TestHub · Accepted Answer

(I) Melting 20 g ice needs 1600 cal; 40 g water gives 2000 cal. $20 (80) + 20 T = 40 (50 - T) \Rightarrow T = 6.6 7^{\circ} C$ (Q).

(II) $50 (80 + T) = 10 (540 + 100 - T) \Rightarrow T = 4 0^{\circ} C$ (R).

(III) Water needs 5000 cal to reach 100°C; only $5000/540 < 10$ g steam condenses. $T = 10 0^{\circ} C$ (T).

(IV) Available heat $5 (540 + 100) + 16 (50) = 4000$ cal melts 50 g of 60 g ice. $T = 0^{\circ} C$ (P).

Matching $I \to Q, II \to R, III \to T, I V \to P$ : C.

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