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Standard X

Physics

Principle of Calorimetry

A mass m1 of ...

Question

A mass $m_{1}$ of a substance of specific heat capacity $c_{1}$ at temperature $t_{1}$ is mixed with a mass $m_{2}$ of other substance of specific heat capacity $c_{2}$ at a lower temperature $t_{2}$ .Deduce the expression for the temperature t of the mixture.State the assumption made,if any.

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Solution

Assuming that no heat is lost to the surrounding;
Heat lost by hot body = Heat gained by cold body.

Given, Heat lost by hot body = $m_{1} \times c_{1} \times (t_{1} - t)$
(because $t_{1}$ is at higher temperature)

And Heat gained by cold body = $m_{2} \times c_{2} \times (t - t_{2})$

Now,
$m_{1} \times c_{1} \times (t_{1} - t)$ = $m_{2} \times c_{2} \times (t - t_{2})$

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Principle of Calorimetry

Standard X Physics

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A mass m1 of a substance of specific heat capacity c1 at temperature t1 is mixed with a mass m2 of other substance of specific heat capacity c2 at a lower temperature t2.Deduce the expression for the temperature t of the mixture.State the assumption made,if any.

Assuming that no heat is lost to the surrounding; Heat lost by hot body = Heat gained by cold body. Given, Heat lost by hot body = m1× c1× (t1−t) (because t1 is at higher temperature) And Heat gained by cold body = m2× c2× (t−t2) Now, m1× c1× (t1−t) =m2× c2× (t−t2)

A mass $m_{1}$ of a substance of specific heat capacity $c_{1}$ at temperature $t_{1}$ is mixed with a mass $m_{2}$ of other substance of specific heat capacity $c_{2}$ at a lower temperature $t_{2}$ .Deduce the expression for the temperature t of the mixture.State the assumption made,if any.