Question

what will be the final temperature is 0.05kg water at 0 degree centigrade is added to 0.25kg pf water at 90 degree centigrade specific heat capacity of water is 4200J/kgk​

Answer 1

Q in = Q out

0.25 x 4200 x (90-t)=0.05 x 4200 x (t-0)

1050 (90-t)=210t

94500-1050t=210t

94500=1260t

t=75°