Question

800 gm water of 80 degree celsius, 10 gm steam of 100 degree Celsius mixed then what is the temperature of mixture​

Answer 1

Answer:

Latent heat of ice L

ice

=336 kJ/kg

Latent heat of steam L

steam

= 2260kJ/kg

Specific heat of water S

water

=4.18 kJ/kg

o

C

Mass of steam, M

steam

=0.01kg

Mass of ice, M

ice

=0.05kg

If final temperature is T

f

Heat Loos from steam = heat gain by ice

M

steam

L

steam

+M

steam

S

water

(100−T

f

)=M

ice

L

ice

+M

ice

S

water

(T

f

−0)

T

f

=

(M

ice

+M

steam

)S

water

M

steam

L

steam

−M

ice

L

ice

+M

steam

S

water

×100

T

f

=

(0.01+0.05)×4.2

0.01×2260−0.05×336+0.01×4.2×100

T

f

=39.6

o

C≅40

o

C

Final temperature of mixture is 40

o

C