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Standard IX

Physics

Relative Density

200g of a cer...

Question

$200 g$ of a certain metal at $83^{\circ} C$ is immersed in $300 g$ of water at $30^{\circ} C$ . The final temperature is $33^{o} C .$ Calculate the specific heat capacity of the metal. Assume that the specific heat capacity of water is $4.2 J g^{- 1} K^{- 1}$

0.378 J g^{- 1} K^{- 1}

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0.578 J g^{- 1} K^{- 1}

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0.778 J g^{- 1} K^{- 1}

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None of the above

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Solution

The correct option is A $0.378 J g^{- 1} K^{- 1}$
Heat loss by metal $=$ Heat gained by water
So, $200 \times c \times (83 - 33) = 300 \times 4.2 \times (33 - 30)$
$\Rightarrow c = 0.378 J / g K$

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Similar questions

200 g

mass of a certain metal at

83^{\circ} C

is immersed in

300 g

of water of

30^{\circ} C

. Final temperature of combination is

33^{o} C

.Calculate the specific heat capacity of metal. Assume that the specific heat capacity of water is

4.2 J g^{- 1} K^{- 1}

A mass of 50 g of a certain metal at $150^{\circ} C$ is immersed in 100 g of water at $11^{\circ} C$ . The final temperature is $20^{\circ} C$ . Calculate the specific heat capacity of the metal. Assume that the specific heat capacity of water is 4.2 J $g^{- 1} K^{- 1}$ .

Q. Calculate the total amount of heat energy required to convert

100 g

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completely into water at

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. Specific heat capacity of ice =

2.1 J g^{- 1} K^{- 1}

, specific heat capacity of water =

4.2 J g^{- 1} K^{- 1}

, specific latent heat of ice =

336 J g^{- 1}

200 g

of hot water at

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of cold water at

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. Neglecting the heat taken by the container, calculate the final temperature of the mixture of water. Specific heat capacity of water =

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A refrigerator converts 100 g of water at $20^{o} C$ to ice at $- 10^{o} C$ in 73.5 min. Calculate the average rate of heat extraction in watt. The specific heat capacity of water is 4.2 J $g^{- 1} K^{- 1}$ , specific latent heat of ice is 336 J $g^{-} 1$ and the specific heat capacity of ice is 2.1 J $g^{- 1} K^{- 1}$ .

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200g of a certain metal at 83∘C is immersed in 300g of water at 30∘C. The final temperature is 33oC. Calculate the specific heat capacity of the metal. Assume that the specific heat capacity of water is 4.2Jg−1K−1

The correct option is A 0.378 Jg−1K−1Heat loss by metal = Heat gained by waterSo, 200×c×(83−33)=300×4.2×(33−30)⇒c=0.378 J/gK

$200 g$ of a certain metal at $83^{\circ} C$ is immersed in $300 g$ of water at $30^{\circ} C$ . The final temperature is $33^{o} C .$ Calculate the specific heat capacity of the metal. Assume that the specific heat capacity of water is $4.2 J g^{- 1} K^{- 1}$

The correct option is A $0.378 J g^{- 1} K^{- 1}$
Heat loss by metal $=$ Heat gained by water
So, $200 \times c \times (83 - 33) = 300 \times 4.2 \times (33 - 30)$
$\Rightarrow c = 0.378 J / g K$