10gm of stram at 100°C is passed into a mixture of 100gm of water and 10gm of ice at 0°C. Find the resulting temperature of the mixture.
Answers
40
o
C
Given,
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
Answer:
Correct option is
C
40
o
C
Given,
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
C
Final temperature of mixture is 40
o
C
Explanation:
ans 40°