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Question

# 20 gm ice at −10 ∘C is mixed with m gm steam at 100 ∘C. The minimum value of m so that finally all ice and steam converts into water is [ Use, specific heat and latent heat as Cice=0.5 cal/gm ∘C,Cwater=1 cal/gm ∘C,Lmelt =80 cal/gm and Lvapor =540 cal/gm ]

A
18527 gm
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B
13517 gm
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C
8532 gm
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D
11317 gm
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Solution

## The correct option is C 8532 gmFor minimum value of m, the final temperature of the mixture must be 0 ∘C. We know that, Heat lost by steam = Heat gained by ice ⇒mLvapor+mCwaterΔT=miceCiceΔT+miceLmelt ⇒m×540+m×1×100=20×0.5×10+20×80 ⇒m=8532 gm

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