$45 g$ 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.

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Solution

Assumption : Assume no. heat loss to surrounding
So,
$\begin{matrix} {(Δ Q)}_{a b s o r b} = {(Δ Q)}_{g i v e n} ∴ T_{F} = \frac{m_{1} s_{1} t_{1} + m_{2} s_{2} t_{2}}{m_{1} s_{1} + m_{2} s_{2}} = \frac{45 \times 4.2 \times 323 + 50 \times 0.39 \times 291}{45 \times 4.2 + 50 \times 0.39} = \frac{66721.5}{208.5} = 320 k a n d 320 k = 320 - 273^{\circ} C = 47^{\circ} C \end{matrix}$

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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 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.

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

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 °C is present 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 °C.

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Specific heat capacity of copper vessel= 400 J kg^-1 °C^-1

Specific heat capacity of water 4200 J kg^-1 °C^-1

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Assumption : Assume no. heat loss to surrounding So,(ΔQ)absorb=(ΔQ)given∴TF=m1s1t1+m2s2t2m1s1+m2s2=45×4.2×323+50×0.39×29145×4.2+50×0.39=66721.5208.5=320kand320k=320−273∘C=47∘C