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A piece of metal of mass 112g is heated to $100^{0} C$ and dropped into a copper calorimeter of mass 40g containing 200g of water at $16^{0} C$ . Neglecting heat loss, if the equilibrium temperature reached is ${24.1}^{0} C$ , then the specific heat of the metal is nearly
(Specific heat of copper = $0.1 cal / {gm}^{0} C$ )

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$0.494 c a l / g m^{0} C$

$0.294 c a l / g m^{0} C$

$0.194 c a l / g m^{0} C$

$0.394 c a l / g m^{0} C$

answer is C.

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Detailed Solution

By the principle of method of mixture,

Heat lost by metal = Heat gained by calorimetry and water

$\Rightarrow 112 \times S_{metal} \times 75.9 = 40 \times 0.1 \times 8 + 200 \times 1 \times 8$

$\Rightarrow S_{metal} = 0.192 \frac{cal}{g {}^{0}C}$

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