Question

what could be the final temperature of a mixture of 100 g of water at 90"c and 600 g of water at 20"c

Answer 1

Let final Temperature = T

Here heat lost by 100 g at 90° = heat gained by 600 g at 20°

Now by the formula,

100 × S × ( 90° - T ) = 600 × S × ( T- 20° )

(90° -T) = [600 × S × (T-20°)] / (100 × S)

⇒ 90° - T = 6(T - 20°)

⇒ 90° - T = 6T - 120

⇒ 120 +90 = 7T

7T = 210

T = 210/7

T = 30°

Hence the final temperature of the mixture is 30°

Hope it Helps

Answer 2

heat lost by water = heat gained by cold water
m*c*t =m2*c*t2
0.1g*4200*(90-t)=0.6*4200*(20-t2)
0.1*(90-t) =0.6*(20-t2)
9-0.1t=12-0.6t2
0.1-0.6=-12-9
-0.7=-21
0.7=21
21*10/0.7*10=210/70=30
final temperature of a mixture =30"C