If 30 g ice at -20°C, 65 g water at 50°C and 5g steam at 100°C are mixed in a container which have negligible heat capacity. [Specific heat of ice = 0.5 cal / g ∘ C , specific heat of water = 1 cal /

Question

If 30 g ice at -20°C, 65 g water at 50°C and 5g steam at 100°C are mixed in a container which have negligible heat capacity. [Specific heat of ice = 0.5 cal / g ∘ C , specific heat of water = 1 cal / g ∘ C , L f ​ = 80 cal / g , L v ​ = 540 cal/g] then :-

TestHub · Accepted Answer

Heat needed to bring 30 g ice at -20°C to water at 0°C: $30 (0.5) (20) + 30 (80) = 2700 cal$ .

Heat released by water and steam cooling to 0°C: $65 (50) + 5 (540) + 5 (100) = 6450 cal > 2700 cal$ , so all ice melts and T > 0°C.

$2700 + 30 T = 65 (50 - T) + 5 (540) + 5 (100 - T) \Rightarrow 100 T = 3750 \Rightarrow T = 37. 5^{\circ} C$ ; total water = 30 + 65 + 5 = 100 g.

Answer: (A), (C)

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