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
Given their respective sp.heat as
By calorimetric principle system has not taken heat from surrounding.
So net heat gain is zero.
Hence