Question

When 30gm of water at 30 degree centigrade is added to 50 degree centigrade the resultant temperature of the solid is.......

Answer 1

  T =  (m1  T1  + m2 T2 ] / (m1 + m2)
     =  (30 * 30 + 50 * 50 ) / (30+50)
     = 3400 / 80
     = 42.5 deg centigrade

Answer 2

Final Temperature T = [M1T1+M2T2] / [M1+M2]
Here,
Given M1 = 30gm, M2 = 50gm, T1 = 30 degree centigrade, T2 = 50 degree centigrade
                           T = [30×30+50×50] / [30+50]
                           T = 900+2500/80
                           T = 3400/80
                           T = 42.5 Degree Centigrade
Therefore, Final Temperature T = 42.5 Degree Centigrade