A liquid P of specific heat capacity 2400 J kg -1 K -1 and at 70 -C is mixed with another liquid R of specific heat capacity 1000 J kg -1 K -1 at 30 -C- After mixing- the final temperature of the mixture is 40 -C- Find the ratio of the mass of the liquids mixed-

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Byju's Answer

Standard X

Physics

Calorimeter

A liquid P of...

Question

A liquid P of specific heat capacity 2400 $J$ ${k g}^{- 1}$ $K^{- 1}$ and at $70^{\circ} C$ is mixed with another liquid R of specific heat capacity 1000 $J$ ${k g}^{- 1}$ $K^{- 1}$ at $30^{\circ} C$ . After mixing, the final temperature of the mixture is $40^{\circ} C$ . Find the ratio of the mass of the liquids mixed?

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Solution

Given : $S_{P} = 2400 J k g^{- 1} K^{- 1}$ and $S_{R} = 1000 J k g^{- 1} K^{- 1}$
As no heat is absorbed or released between the system and surrounding.
Thus sum of the heat absorbed between the liquids is zero.
$∴$ $M_{p} S_{P} (Δ T_{P}) + M_{R} S_{R} (Δ T_{R}) = 0$
Or $M_{P} (2400) (40 - 70) + M_{R} (1000) (40 - 30) = 0$
Or $M_{P} (2400) (30) = M_{R} (1000) (10)$
We get $\frac{M_{P}}{M_{R}} = \frac{(1000) (10)}{(2400) (30)} = \frac{5}{36}$

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A liquid P of specific heat capacity 2400 J kg−1 K−1 and at 70∘C is mixed with another liquid R of specific heat capacity 1000 J kg−1 K−1 at 30∘C. After mixing, the final temperature of the mixture is 40∘C. Find the ratio of the mass of the liquids mixed?