Question

if 40 gram of water at 100 degree celsius is mixed with 150 gram of water and mixture temperature becomes 50 degree celcius. Find initial temperature of cold water.

Answer 1

we have a formula T=(mt1+m2t2)/(m1+m2)

here m1=mass of water=40g

         t1=temperature of water=100deg c

          m2=  mass of second water taken=150

          t2=temperature of second water taken

          T=50

50=40(100)+150(t2)/(40+150)

50=(4000+150t2)/190

9500=4000+150t2

9500-4000=150t2

5500=150t2

5500/150=t2 

36.6=t2

therefore the intial temperature of cold water is 36.6deg celcius

Answer 2

let s be the specific heat capacity of water. Let the temperature of cold water be T deg Celsius.

Heat lost by hot water = heat gained by cold water
m1 s ΔT = m2 s ΔT

40gm  * s Cal/degK /gm * (100 - 50) degK = 150gm * s cal/gm/degK * (50 - T)
40 * 50 = 150 (50-T)
2000 = 7500 - 150 T

150 T = 5500

T = 36.666 deg Celsius