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Principle of Calorimetry

45 g of water...

Question

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 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.

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Solution

$M a s s space o f space w a t e r space left parenthesis m subscript 1 right parenthesis equals 45 g t e m p e r a t u r e space o f space w a t e r space left parenthesis T subscript 1 right parenthesis equals 50 degree M a s s space o f space c o p p e r space left parenthesis m subscript 2 right parenthesis equals 50 g t e m p e r a t u r e space o f space c o p p e r space left parenthesis T subscript 2 right parenthesis equals 18 degree C F i n a l space t e m p e r a t u r e space left parenthesis T right parenthesis equals ? T h e space s p e c i f i c space h e a t space c a p a c i t y space o f space t h e space c o p p e r space equals 0.39 space J divided by g divided by space K T h e space s p e c i f i c space h e a t space c a p a c i t y space o f space w a t e r space C subscript 1 equals 4.2 space J divided by g divided by K m subscript 1 c subscript 1 left parenthesis T subscript 1 minus T right parenthesis equals m subscript 2 c subscript 2 left parenthesis T minus T subscript 2 right parenthesis T equals fraction numerator m subscript 1 c subscript 1 T subscript 1 plus end subscript m subscript 2 c subscript 2 T subscript 2 over denominator m subscript 1 c subscript 1 plus m subscript 2 c subscript 2 end fraction space equals fraction numerator 45 cross times 4.2 cross times 50 plus 50 cross times 0.39 cross times 18 over denominator 45 cross times 4.2 plus 50 cross times 0.39 end fraction equals fraction numerator 9801 over denominator 208.5 end fraction equals 47 degree C T equals 47 degree C$

Here, we have made the assumption that there is no loss of heat.

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Similar questions

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

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.

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

. Find the specific heat capacity of the metal.
Specific heat capacity of water is

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

250 g

of water at

30^{\circ} 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^{\circ} 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}

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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Principle of Calorimetry

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