Question

Two liquids A and B are at 32 degree Celsius and 24 degree Celsius and makes it equal masses the temperature of the mixture is found to be 28 degree Celsius the specific heats are in the ratio of

Answer 1

answer is 1:1

Explanation
temperature of mixture Φmix = ΦA cA + ΦB cB/cA+cB
=> 28= 32×cA+24×cB/cA+cB
=> 29 cA+ 28 cB=32cA+24cB
=> cA/cB=1/1

Answer 2

Answer:

The ratio of specific heats of the liquids is 1:1.

Explanation:

From the principle of calorimetry, we know,

                     Heat lost = Heat gained    (1)

So,

     Heat lost by liquid A=Heat lost by liquid B

                  $Q_A=Q_B$    

       $m_AS_A(T_1-T_0)=m_BS_B(T_0-T_2)$    (2)

Where,

$m_A,m_B$=respective masses of the liquids

$S_A$=specific heat capacity of liquid A

$S_B$=specific heat capacity of liquid B

T₁=temperature at which liquid A is present

T₂=temperature at which liquid B is present

T₀=temperature of the mixture

From the question we have,

The masses of liquids are equal.

T₁=32°C

T₂=24°C

T₀=28°C

By putting the values in equation (2) we get;

$mS_A(32\textdegree C-28\textdegree C)=m_BS_B(28\textdegree C-24\textdegree C)$

$S_A\times 4=S_B\times 4$

$\frac{S_A}{S_B} =\frac{1}{1}$

Hence, the ratio of specific heats of the liquids is 1:1.

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