A piece of iron of mass 0.05 kg is heated to a temperature of $200^{o} C$ and dropped into a beaker containing 0.4 kg of water at $20^{o} C$ . If the final temperature of water and iron is ${22.4}^{o} C$ , find the specific heat capacity of iron.

354.04 J k g^{- 1} K^{- 1}

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454.04 J k g^{- 1} K^{- 1}

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554.04 J k g^{- 1} K^{- 1}

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654.04 J k g^{- 1} K^{- 1}

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Solution

The correct option is B $454.04 J k g^{- 1} K^{- 1}$
Mass of iron, $M = 0.05 k g$
Initial temperature of iron, $t_{1} = 200^{o} C$
Final temperature of iron, $t_{2} = {22.4}^{o} C$
Mass of water, $m = 0.4 k g$
Initial temperature of water, $t_{3} = 20^{o} C$
Final temperature of water, $t_{2} = {22.4}^{o} C$
Specific heat capacity of water
$= 4200 J k g^{- 1} K^{- 1}$
Let specific heat capacity of iron $= C$
Heat lose by iron $= M C (t_{1} - t_{2})$
$= 0.05 \times C \times (200 - 22.4)$
$= 0.05 \times C \times 177.6 = 8.88 C J$
Heat gained by water
$= m \times 4200 \times (t_{2} - t_{3})$
$= 0.4 \times 4200 \times (22.4 - 20)$
$= 0.4 \times 4200 \times 2.4 = 4032 J$
Equating heat lost and heat gained
$8.88 C = 4032$
Therefore, $C = 4032 / 8.88$
$= 454.04 J k g^{- 1} K^{- 1}$ .

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Similar questions

Q. A piece of iron of mass

0.05 k g

is heated to a temperature of

200^{\circ} C

and dropped into a beaker containing

0.4 k g

of water at

20^{\circ} C

. If the final temperature of water and iron is

{22.4}^{\circ} C

, find the specific heat capacity of iron.

An aluminium vessel of mass 0.5 kg contains 0.2 kg of water at $20^{\circ} C$ . A block of iron of mass 0.2 kg at $100^{\circ} C$ is gently put into the water. Find the equilibrium temperature of the mixture. Specific heat capacities of aluminium, iron and water are $910 K g^{- 1} K^{- 1} 470 J k g^{- 1} K^{- 1}$ and $4200 J k g^{- 1} K^{- 1}$ respectively.