What will be the value of g for ice at 8c temperature?
Answers
Answer:
Given: mass of ice, m_{i}=50gm
i
=50g
mass of water, m_{w }=50gm
w
=50g
temperature of ice, T_{i}=0^{o}CT
i
=0
o
C
temperature of water, T_{w}=100^{o}CT
w
=100
o
C
We have,
specific heat of water, c_{w}=1\ cal/g/^{o}Cc
w
=1 cal/g/
o
C
latent heat of ice, L_{w}=89\ cal/gL
w
=89 cal/g
Let T^{o}CT
o
C be the final temperature of mixture.
Now energy balance equation:
Heat gained by ice to change itself into water ++ heat gained by melted ice(water) to raise its temperature at T^{o}C=T
o
C= heat lost by water to reach at T^{o}CT
o
C
m_{i}L_{i}+m_{i}c_{w}(T-0)=m_{w}c_{w}(100-T)m
i
L
i
+m
i
c
w
(T−0)=m
w
c
w
(100−T)
50\times80+50\times1(T-0)=50\times1(100-T)50×80+50×1(T−0)=50×1(100−T)
4000+50T=5000-50T4000+50T=5000−50T
100T=5000-4000=1000100T=5000−4000=1000
T=\dfrac{1000}{100}=10^{o}CT=
100
1000
=10
o
C