Question

Two liquids of masses m, 2m and specific heats 3s, 2s are at temperature 20°C, 40°C
respectively. Find resultant temperature when they are mixed

Answer 1

Given :

Masses of two liquids = m and 2m

Specific heat = 3s and 2s respectively

Temperature of two liquids = 20° C and 40°C

To find :

The resultant temperature of mixture T

Solution :

It is given that

mass of liquid A = m

mass of liquid B = 2m

Specific heat of liquid A = 3s

Specific heat of liquid B = 2s

Temperature of liquid A = 20°C

Temperature of liquid B = 40°C

Heat in a system remains constant so, in this reaction when two liquids are mixed together the heat dissipated by one liquid is gained by other liquid.

On mixing both liquids

Let the temperature of mixture is T ,it means

temperature of liquid A increase to T and temperature of liquid B decrease to T

So the heat dissipated by liquid B in decreasing its temperature is gained by liquid A to increase its temperature.

So

Heat gained by liquid A = Heat lost by liquid B

$M_A\times S_A (T-20) = M_B\times S_B (40-T) \\ \\ m \times 3s \times (T-20) = 2m \times 2s \times (40-T) \: \:$

So,

3ms(T-20) = 4ms(40-T)

On cancelling ms from both sides,

3T - 60 = 160 - 4T

3T - 60 = 160 - 4T

7T = 220°C

so

$\textbf{ \Large \: Temperature of \: mixture T = } \frac{ \textbf{ \Large \: 220}}{ \textbf{ \Large \: 7}}$

So, temperature of mixture = 31.43°C

Answer 2

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