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# A liquid of mass m and specific heat s is at temperature $\theta$. Another liquid of mass 2m and specific heat s/2 is at $3\theta$. If these two liquids are mixed, then calculate the resultant temperature of the mixture.

Last updated date: 25th Jun 2024
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Hint:Use the formula for the heat gained or lost by the substance. The heat gained by the liquid of mass m is equal to the heat lost by the liquid of mass 2m.

Formula used:
$\Rightarrow Q = mc\Delta T$
Here, m is the mass of the liquid, c is the specific heat of the liquid and $\Delta T$ is the difference in temperature.

When mixing the two liquids kept at different temperatures, the resultant temperature of the mixture becomes greater than the temperature of cold liquid and less than the temperature of the hot liquid because, the liquid kept at more temperature loses heat and the liquid kept at less temperature gains the heat.
Thus, the heat gained by the liquid of mass m is equal to the heat lost by the liquid of mass 2m.
Let the final temperature of the mixture is T.
The expression for the heat gained by the liquid or heat lost by the liquid is,
$\Rightarrow Q = mc\Delta T$
Here, m is the mass of the liquid, c is the specific heat of the liquid and $\Delta T$ is the difference in temperature.
Since the heat gained by the liquid of mass m is equal to the heat lost by the liquid of mass 2m, we can write,
$\Rightarrow ms\left( {T - \theta } \right) = \left( {2m} \right)\left( {\dfrac{s}{2}} \right)\left( {3\theta - T} \right)$
$\Rightarrow T - \theta = 3\theta - T$
$\Rightarrow 2T = 4\theta$
$\Rightarrow\therefore T = 2\theta$

Therefore, the temperature of the mixture is $2\theta$.

Note:The difference in the temperature of the liquid and final temperature of the mixture should be taken as positive value. For this, subtract the less temperature from the more temperature of the mixture or the liquid.