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Standard XII

Physics

Calorimetry

The water equ...

Question

The water equivalent of a calorimeter is $10$ g and it contains $50$ g of water at $15^{o}$ C. Some amount of ice, initially at $- 10^{o}$ C is dropped in it and half of the ice melts till equilibrium is reached. What was the initial amount of ice that was dropped (when specific heat of ice $= 0.5$ cal $g m^{- 1}$ ${^{o} C}^{- 1}$ , specific heat of water $= 1.0$ cal $g m^{- 1} {^{o} C}^{- 1}$ and latest heat of melting of ice $= 80$ cal $g m^{- 1}$ )?

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Solution

The correct option is C $20$ g
$\Rightarrow m \times \frac{1}{2} \times 10 + \frac{m}{2} \times 80 = 60 \times 1 \times 15$
$\Rightarrow m = \frac{60 \times 15}{45} = 20$ gm.

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Similar questions

Q. 5 g of water at

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and 5 g of ice at

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are mixed together in a calorimeter. What is the final temperature of the mixture. Given specific heat of ice

= 0.5 c a l g^{- 1} (^{o} C)^{- 1}

and latent heat of fusion of ice

= 80 c a l g^{- 1}

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The correct option is C 20g⇒m×12×10+m2×80=60×1×15⇒m=60×1545=20gm.

The correct option is C $20$ g
$\Rightarrow m \times \frac{1}{2} \times 10 + \frac{m}{2} \times 80 = 60 \times 1 \times 15$
$\Rightarrow m = \frac{60 \times 15}{45} = 20$ gm.