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
Jun 29, 2017

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

31.4"^@C, rounded to one decimal place

Explanation:

Let us mix liquids a and b first.
Suppose their final temperature is T^@C

Using Law of Conservation of energy
Heat gained by a= Heat Lost by b

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

Dividing both sides by m, inserting given values of specific heats, we get
1xx(T-20)=0.5xx(40-T)
Multiplying both sides with 2 we get

2T-40=40-T
=>3T=80
=>T=80/3 "^@C

Let us add liquid c to this mixture. Let the equilibrium temperature of this mixture be T_3

Using Law of Conservation of energy again
Heat gained by mixture of aand b= Heat Lost by c

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

Dividing by m and inserting given values of specific heats

[1xx(T_3-80/3)+0.5xx(T_3-80/3)]=0.25xx(60-T_3)

Multiply both sides with 4 and simplify

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

Jun 29, 2017

Let the equilibrium temperature of the mixture of three liquid (a,b.c) each of same mass mg be t^@C. ( assuming cgs system for calculation)
Given their respective sp.heat as s_a=1,s_b=0.5 and s_c=0.25
By calorimetric principle system has not taken heat from surrounding.
So net heat gain is zero.
Hence

mxxs_axx(t-20)+mxxs_bxx(t-40)+mxxs_cxx(t-60)=0

=>s_axx(t-20)+s_bxx(t-40)+s_cxx(t-60)=0

=>1xx(t-20)+0.5xx(t-40)+0.25xx(t-60)=0

=>t-20+0.5t-20+0.25t-15=0

=>1.75t=55

=>t=55/1.75=220/7~~31.4^@C