Question

0.2 m^3 of water at 80 degree c is mixed with 0.6m^3 of water at 20 degree c. find mixture temperature

Answer 1

Answer:

Let the final temperature of the mixture be t,

Heat lost by water at 80

0

C

msΔt

=0.1×10

3

×s

water

×(80

0

−t)

(∵m=V×d=0.1×10

3

kg)

Heat gained by water at 60

0

C

=0.3×10

3

×S

water

×(t−60

0

)

According to principle of calometry

Heat lost =Heat gained

∴0.1×10

3

×S

water

×(80

0

−t)=0.3×10

3

×s

water

×(t−60

0

)

or (80

0

−t)=3×(t−60

0

)

or 4t=260

0

or t=65

0