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Standard X

Physics

Calorimeter

A 50 g piec...

Question

A $50$ g piece of iron at $100^{o}$ C is dropped into $100$ g water at $20^{o}$ C. The temperature of the mixture is ${25.5}^{o}$ C. The specific heat of iron ( in cal $g^{- 1}$ $^{o} C^{- 1}$ ) is :

0.148

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0.082

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0.267

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0.341

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Solution

The correct option is A 0.148
Heat lost by iron =Heat gained by water
Let the specific heat of iron be $c$
$50 \times c \times (100 - 25.5) = 100 \times 1 \times (25.5 - 20)$
on solving we get $c = 0.147 c a l g^{- 1} C^{- 1}$

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A 50 g piece of iron at 100o C is dropped into 100 g water at 20o C. The temperature of the mixture is 25.5o C. The specific heat of iron ( in cal g−1 oC−1) is :

The correct option is A 0.148Heat lost by iron =Heat gained by waterLet the specific heat of iron be c50×c×(100−25.5)=100×1×(25.5−20)on solving we get c=0.147 cal g−1C−1

A $50$ g piece of iron at $100^{o}$ C is dropped into $100$ g water at $20^{o}$ C. The temperature of the mixture is ${25.5}^{o}$ C. The specific heat of iron ( in cal $g^{- 1}$ $^{o} C^{- 1}$ ) is :

The correct option is A 0.148
Heat lost by iron =Heat gained by water
Let the specific heat of iron be $c$
$50 \times c \times (100 - 25.5) = 100 \times 1 \times (25.5 - 20)$
on solving we get $c = 0.147 c a l g^{- 1} C^{- 1}$