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Byju's Answer

Standard IX

Chemistry

Effect of Change of Pressure on State Change of Matter

45 g of water...

Question

$45 g$ of water at $50^{o} C$ in a breaker is cooled when $50 g$ of copper at $18^{o} C$ is added to it. The contents are stirred till a final constant temperature is reached. Calculate this final temperature. The specific heat capacity of copper is $0.39 J g^{- 1} K^{- 1}$ and that of water is $4.2 J g^{- 1} K^{- 1}$ . State the assumption used

47^{o} C

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30^{o} C

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57^{o} C

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None of the above

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Solution

The correct option is A $47^{o} C$
Heat loss by water $=$ Heat gain by copper
So, $45 \times 4.2 \times (50 - t) = 50 \times 0.39 \times (t - 18)$
$\Rightarrow t = 47^{o} C$

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Similar questions

45 g

of water at

50^{\circ} C

in a beaker is cooled when

50 g

of copper at

18^{\circ} C

is added to it. The contents are stirred till a final constant temperature is reached. Calculate this final temperature. The specific heat capacity of copper is

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

and that of water is

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

45 g

of water at

50^{\circ} C

in a beaker is cooled when

50 g

of copper at

18^{\circ} C

is added to it. The contents are stirred till final constant temperature is reached. Calculate the final temperature. The specific heat capacity of copper is

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

and that of water is

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

. State the assumption used.

250 g

of water at

30^{o} C

is contained in a copper vessel of mass

50 g

. Calculate the mass of ice required to bring down the temperature of the vessel and its contents to

5^{o} C

. Given : specific latent heat of fusion of ice

= 336 \times 10^{3} J k g^{- 1}

, specific heat capacity of copper

= 400 J k g^{- 1} K^{- 1}

, specific heat capacity of water

= 4200 J k g^{- 1} K^{- 1}

Q. In an experiment,

17 g

of ice is used to bring down the temperature of

40 g

of water at

34^{0} C

to its freezing temperature. The specific heat capacity of water is

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

. Calculate the specific latent heat of ice. State one important assumption made in the above calculation.

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 $g^{- 1} K^{- 1}$ .

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The correct option is A 47oCHeat loss by water = Heat gain by copperSo, 45×4.2×(50−t)=50×0.39×(t−18)⇒t=47oC

The correct option is A $47^{o} C$
Heat loss by water $=$ Heat gain by copper
So, $45 \times 4.2 \times (50 - t) = 50 \times 0.39 \times (t - 18)$
$\Rightarrow t = 47^{o} C$