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Q.

5 gm of steam at 100°C is passed into calorimeter containing liquid. Temperature of liquid rises from 32°C to 40°C. Then water equivalent of calorimeter and contents is

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a

160 gram

b

300 gram

c

40 gram

d

375 gram

answer is B.

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Detailed Solution

msLs+msst2−t=mwst−t1
⇒5(540)+5(1)(100−40)=mw(40−32)
⇒mw=375 gm

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