Question

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?

Answer 1

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

Answer 2

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`