20g of steam at $100 ° C$ is passed into 100g of ice at $0 ° C$ . Find the resultant temperature (Latent heat of steam is 540 cal/g and $L_{i c e} = 80 c a l / g;$ $S_{W} = 1 c a l / g ° C$ )

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$100^{\circ}$

$60^{\circ}$

$80^{\circ}$

$40^{\circ}$

answer is B.

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Detailed Solution

Heat lost by steam = Heat gained by ice

m₁L₁+ m₁S(T₁- T₃) = m₂L₂+ m₂S(T₃- T₂)

From given data,

20×540 + 20×1×(100 - T₃) = 100×80 + 100×1×(T₃- 0)

10800 + 2000 - 20T₃ = 8000 + 100T₃

12800 - 8000 = 120T₃

T₃ = 40°C

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