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 #m#g 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#