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Standard XII

Physics

Calorimetry

200 g of hot ...

Question

$200 g$ of hot water at $80^{\circ} C$ is added to $300 g$ of cold water at $10^{\circ} C$ . Neglecting the heat taken by the container, calculate the final temperature of the mixture of water. Specific heat capacity of water = $44200 J k g^{- 1} K^{- 1}$ .

38 K

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19^{\circ} C

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38^{\circ} C

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none of the above

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Solution

The correct option is B $38^{\circ} C$
The final temperature is given as $\frac{m_{1} t_{1} + m_{2} t_{2}}{m_{1} + m_{2}} = \frac{200 \times 80 + 300 \times 10}{200 + 300} = 19000 / 500 = 38^{o} C$

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Similar questions

200 g

of hot water at

80^{o} C

is added to

300 g

of cold water at

10^{o} C

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

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

200 g of hot water at 80°C is added to 300 g of cold water at 10°C. Calculate the final temperature of the mixture of water. Consider the heat taken by the container to be negligible. The specific heat capacity of water is 4200 J kg^-1 °C^-1

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Q. A piece of ice of mass

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200 g of hot water at 80∘C is added to 300 g of cold water at 10∘C. Neglecting the heat taken by the container, calculate the final temperature of the mixture of water. Specific heat capacity of water = 44200 Jkg−1K−1.

The correct option is B 38∘CThe final temperature is given as m1t1+m2t2m1+m2=200×80+300×10200+300=19000/500=38oC

$200 g$ of hot water at $80^{\circ} C$ is added to $300 g$ of cold water at $10^{\circ} C$ . Neglecting the heat taken by the container, calculate the final temperature of the mixture of water. Specific heat capacity of water = $44200 J k g^{- 1} K^{- 1}$ .

The correct option is B $38^{\circ} C$
The final temperature is given as $\frac{m_{1} t_{1} + m_{2} t_{2}}{m_{1} + m_{2}} = \frac{200 \times 80 + 300 \times 10}{200 + 300} = 19000 / 500 = 38^{o} C$