Question

Calculate the equilibrium temperature of the mixture formed by mixing 100 ml of water at 50C with 200 ml of water at 800C​

Answer 1

Explanation:

since it is given that 50 gms water at 20 ⁰ C and 50 gms of water at 40⁰ C are mixed.  Since the masses of the liquid at different temperatures are same, the answer is very easy and simple :  average of 20⁰C and 40⁰C.  that is:  30⁰C.

=====================

final temperature of the mixture = 

     = [ m1 * T1 + m2 * T2 ] / (m1 + m2)

     = [ 50 gms * 20⁰ C + 50 gms * 40⁰C ]  / (50+50)

     = 3,000 / 100 = 30⁰C

====================

another way using specific heats :

   let the final temperature be = T ⁰C

   Amount of heat given out by the hot water = m * s * (40⁰C - T)

           = 50 gms * s* (40 -T)

   Amount of heat taken in by the cold water = m * s * (T - 20⁰C)

       = 50 gms * s * (T - 20 )

   As the amounts are equal, because the heat is transferred from hotwater to the cold water without any loss of heat to any surroundings,

           50 * s * (40 -T) = 50 gm * s  * (T-20) 

                 40 - T = T - 20

             2 T = 60    => T = 30⁰C

i hope my answer is correct any doubt you can ask me . thank you.