Question

how much will the temperature of 100g of water be raised by doing 4200J of work is stirring the water?

Answer 1

ANSWER

Let mass of steam required =m gm

each gram of steam of condensing release 536 calories of heat steam condense at 100

o

C

cools finally to 90

o

C

Heat released by m gm of steam on condensing =536×m calne

Final Temp of solution =m× specific heat of water × fall of temp

=m×1×(100−90)

=m×1×10

=10 m clone

heat released =536 m+10 m

546 m calories of heat

Heat required to raised the temp of 100 gm of water at 24

o

C+m gm of condensed steam from 24

o

C−90

o

C

=(100+m)×1×(90−24)

=(100+m)×66 calories

heat gained = Heat lost

(100+m)×66=546 m

6600+66 m=546 m

600=486 m

m=13.75 g of steam

≃13 gm

hope it will be helpful....