Two liquids $L_{1}$ and $L_{2}$ are at $20^{\circ} C$ and $40^{\circ} C$ respectively. When equal masses of them are mixed together the temperature of the mixture becomes $35^{\circ} C$ . The ratio of specific heat capacities of $L_{1}$ and $L_{2}$ is:

1 : 1

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3 : 1

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1 : 3

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2 : 3

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Solution

The correct option is C $1 : 3$
Given: Two liquids $L_{1}$ and $L_{2}$ are at $20^{\circ} C$ and $40^{\circ} C$ respectively. When equal masses of them are mixed together the temperature of the mixture becomes $35^{\circ} C$ .
To find the ratio of specific heat capacities of $L_{1}$ and $L_{2}$
Solution:
The temperature of the mixture can be found out using the formula,
$T = \frac{T_{1} c_{1} + T_{2} c_{2}}{c_{1} + c_{2}}$
where $T = 35^{\circ} C$ is the final temperature of the mixture.
$T_{1} = 20^{\circ} C$ is the temperature of the liquid $L_{1}$
$T_{2} = 40^{\circ}$ is the temperature of the liquid $L_{2}$
$c_{1}$ is the specific heat of liquid $L_{1}$
$c_{2}$ is the specific heat of liquid $L_{2}$
So above equation becomes
$35 = \frac{20 c_{1} + 40 T_{2}}{c_{1} + c_{2}} ⟹ 35 c_{1} + 35 c_{2} = 20 c_{1} + 40 c_{2} ⟹ 35 c_{1} - 20 c_{1} = 40 c_{2} - 35_{2} ⟹ 15 c_{1} = 5 c_{2} ⟹ \frac{c_{1}}{c_{2}} = \frac{5}{15} ⟹ \frac{c_{1}}{c_{2}} = \frac{1}{3}$
is the required ratio of specific heat capacities of $L_{1}$ and $L_{2}$

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