# Three liquids a,b,c having specific heat 1, 0.5 , 0.25 respectively are at temperature 20 ,40 ,60 respectively find temperature equilibrium if all have the same mass?

##### 2 Answers

See the other solution which is shorter and has correct steps.

#### Explanation:

Let us mix liquids

Suppose their final temperature is

Using Law of Conservation of energy

Heat gained by

#ms_a(T-20)=ms_b(40-T)#

Dividing both sides by

Multiplying both sides with

#2T-40=40-T#

#=>3T=80#

#=>T=80/3 "^@C#

Let us add liquid

Using Law of Conservation of energy again

Heat gained by mixture of

#[ms_a(T_3-80/3)+ms_b(T_3-80/3)]=ms_c(60-T_3)#

Dividing by

Multiply both sides with

#6xx(T_3-80/3)=(60-T_3)#

#=>6T_3+T_3=60+160#

#=>T_3=220/7=31.4"^@C# , rounded to one decimal place

Let the equilibrium temperature of the mixture of three liquid (a,b.c) each of same mass *assuming cgs system for calculation*)

Given their respective sp.heat as

By calorimetric principle system has not taken heat from surrounding.

**So net heat gain is zero.**

Hence