Question

Two liquids A and B are at 300C and 200C respectively. When they are mixed in equal masses the temperature of the mixture is found to be 260C. the ratio of specific heats is
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Answer 1

Answer:

Loss in temperature of one liquid is equal to the gain in temperature of another liquid.

m1 .C1 .ΔT1 =m 2 .C 2 .ΔT 2

Given the masses are same

C 1 .(30−26)=C 2 .(26−20)c 2

C1/C2C = 6/4

=3/2

Explanation:

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Answer 2

Answer:

The ratio of specific heats is 3:2

Explanation:

Given that,

The temperature of liquid A=300°C

The temperature of liquid B=200°C

The temperature of Mixture=260°C

Let the mass of each liquid=m

According to thermal equilibrium,

Heat lost by liquid A=Heat gained by liquid B

$(Q _{A}) _{lost} = (Q _{B} )\: _{gain}$

Therefore,based on known values we write the above relation as

$mc _{1}(300 - 260) =mc _{2}(260 - 200) \\ = > 40c_{1} = 60c_{2} \\ \frac{c _{1} }{c_{2} } = \frac{60}{40} = \frac{3}{2}$

Therefore,the ratio of specific heats is 3:2

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