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Standard X

Physics

Specific Heat Capacity

Two liquids a...

Question

Two liquids are temperature $20^{\circ} C$ and $40^{\circ} C$ . When same mass of both of them is mixed, the temperature of the mixture is $32^{\circ} C$ . What is the ratio of their specific heats ?

\frac{1}{3}

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\frac{2}{5}

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\frac{3}{5}

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\frac{2}{3}

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Solution

The correct option is D $\frac{2}{3}$
Final temperatue is $32^{o} C$ .
As $Q = m c Δ T$ and
heat lost by warm liquid=heat gained by cold liquid.
$m_{1} c_{1} (32 - 20) = m_{2} c_{2} (40 - 32)$
given that, $m_{1} = m_{2}$
so, $\frac{c_{1}}{c_{2}} = \frac{2}{3}$

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Two liquids are temperature 20∘C and 40∘C. When same mass of both of them is mixed, the temperature of the mixture is 32∘C . What is the ratio of their specific heats ?

The correct option is D 23Final temperatue is 32oC.As Q=mcΔT andheat lost by warm liquid=heat gained by cold liquid.m1c1(32−20)=m2c2(40−32) given that, m1=m2so, c1c2=23

Two liquids are temperature $20^{\circ} C$ and $40^{\circ} C$ . When same mass of both of them is mixed, the temperature of the mixture is $32^{\circ} C$ . What is the ratio of their specific heats ?

The correct option is D $\frac{2}{3}$
Final temperatue is $32^{o} C$ .
As $Q = m c Δ T$ and
heat lost by warm liquid=heat gained by cold liquid.
$m_{1} c_{1} (32 - 20) = m_{2} c_{2} (40 - 32)$
given that, $m_{1} = m_{2}$
so, $\frac{c_{1}}{c_{2}} = \frac{2}{3}$