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Standard X

Physics

Principle of 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 =4200 J k $g^{- 1} K^{- 1}$ .

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Solution

Let the final temperature be $T^{o} C$ .

So, heat lost by hot water = Heat gained by cold water

$\Rightarrow 0.2 \times 4200 \times (80 - T) = 0.3 \times 4200 \times (T - 10)$

$\Rightarrow 16 - 0.2 T = 0.3 T - 3$

$\Rightarrow 0.5 T = 19$

$∴ T = 38^{o} C$

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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

Q. A piece of ice of mass

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0^{o} C

is added to

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

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and specific latent heat of fusion of ice =

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

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of a certain metal at

83^{\circ} C

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. The final temperature is

33^{o} C .

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

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mass of a certain metal at

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is immersed in

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Principle of Calorimetry

Standard X Physics

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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 =4200 J kg−1 K−1 .

Let the final temperature be ToC. So, heat lost by hot water = Heat gained by cold water ⇒0.2×4200×(80−T)=0.3×4200×(T−10) ⇒16−0.2T=0.3T−3 ⇒0.5T=19 ∴T=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 =4200 J k $g^{- 1} K^{- 1}$ .

Let the final temperature be $T^{o} C$ .

So, heat lost by hot water = Heat gained by cold water

$\Rightarrow 0.2 \times 4200 \times (80 - T) = 0.3 \times 4200 \times (T - 10)$

$\Rightarrow 16 - 0.2 T = 0.3 T - 3$

$\Rightarrow 0.5 T = 19$

$∴ T = 38^{o} C$