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Standard XII

Physics

Area and Volume Expansion

200 g of hot ...

Question

$200 g$ 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 $= 4200 J k g^{- 1} K^{- 1}$

48^{o}

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18^{o}

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28^{o}

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38^{o}

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Solution

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

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Similar questions

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}

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

200 g

of a certain metal at

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is immersed in

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

Q. A piece of ice of mass

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is added to

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of water at

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. Calculate the final temperature of water when all the ice has melted. Specific heat capacity of water =

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

and specific latent heat of fusion of ice =

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200 g

mass of a certain metal at

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is immersed in

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of water of

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200 g of hot water at 80oC is added to 300 g of cold water at 10oC. Neglecting the heat taken by the container, calculate the final temperature of the mixture of water. Specific heat capacity of water =4200Jkg−1K−1

The correct option is D 38o CThe final temperature is given as, m1t1+m2t2(m1+m2)=200×80+300×10(200+300)=19000500=38oC

$200 g$ 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 $= 4200 J k g^{- 1} K^{- 1}$

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