A calorimeter of water equivalent 1 kg contains 10 kg of ice & 10 kg of water in thermal equilibrium. The atmospheric temperature is 1 5 ∘ below freezing point due to which the calorimeter loses heat.

Question

A calorimeter of water equivalent 1 kg contains 10 kg of ice & 10 kg of water in thermal equilibrium. The atmospheric temperature is 1 5 ∘ below freezing point due to which the calorimeter loses heat. As a result ice is formed inside the calorimeter at a rate of 10.8 gm per second. To try to compensate for this heat loss, steam at 10 0 ∘ C is supplied to the calorimeter at a rate of r . ( L V ​ = 540 cal/gm, L f ​ = 80 cal/gm, sp heat of water 1 cal/gm ∘ C .) Column-I gives the value of r and column-II gives the situation just after the introduction of steam. Column I Column II (A) r = 1.6 g/s (P) Amount of ice in calorimeter increases. (B) r = 1.35 g/s (Q) Amount of water in calorimeter increases. (C) r = 1.2 g/s (R) Amount of ice remains constant at 10 kg . (D) r = 1 g/s (S) Amount of water remains constant at 10 kg . (T) Amount of ice in calorimeter decreases.

TestHub · Accepted Answer

Heat loss rate: 80 × 10.8 = 864 cal/s Each gram of steam supplies: 540 + 100 = 640 cal/g At the four steam rates: 1.6 × 640 = 1024 > 864 ( A ) : Q , T 1.35 × 640 = 864 ( B ) : Q , R 1.2 × 640 = 768 < 864 ( C ) : P , S 1 × 640 = 640 < 864 ( D ) : P In (C), freezing balances added steam.

Column I	Column II
(A) $r = 1.6 g/s$	(P) Amount of ice in calorimeter increases.
(B) $r = 1.35 g/s$	(Q) Amount of water in calorimeter increases.
(C) $r = 1.2 g/s$	(R) Amount of ice remains constant at $10 kg$ .
(D) $r = 1 g/s$	(S) Amount of water remains constant at $10 kg$ .
	(T) Amount of ice in calorimeter decreases.

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