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Q.

10 gm of ice at – 100 C is mixed with 100 gm of water at 500 C contained in a calorimeter weighing 50gm. (Specific heat of water = 1 cal. gm–1. 0C–1, Latent heat of ice=80cal/gm, specific heat of ice = 0.5 cal/gm / 0C and specific heat of copper = 0.09 cal/gm/0C). The final temperature reached by the mixtures is

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a

38.20 C

b

40.00 C

c

25.50C

d

30.0 0C

answer is C.

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Detailed Solution

Heat gained by ice
= Heat lost by water + Heat lost by calorimeter
msdT+mL+msdTice=(msdT)w+(msdT)c[(10)(0.5(0(10)]+(10)(80)+(10)(1)(T0) =(100)(1)(50T)+50(0.09)(50T)T=38.2C

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