a liquid of mass 240 gram contained in a calorie metre of water equivalent 12 gram at temperature for 14 degree Celsius A copper piece of mass 72 gram and specific heat 0.093 increased to 100 degree Celsius and then put into calorie metre if final temperature 18 degree Celsius find specific heat of liquid.
Answers
Answer:
Given: A mixture of 250g of water and 200g of ice is kept in a calorimeter of water equivalent 50g. If 200g of steam at 100
o
C is passed through this mixture, Latent heat of steam is 540cal/g.
To find the final temperature and weight of contents of the calorimeter.
Solution:
As per given criteria,
Mass of water, m
w
=250g
Mass of ice, m
i
=200g
Water equivalent, w
e
=50g
Mass of steam, m
s
=200g
Temperature of the steam, T
s
=100
∘
C
Latent heat of steam is L=540cal/g.
And we know,
Specific heat of water, s=1cal/(g
∘
C)
Latent heat of ice, L
i
=80cal/g
According to the principle of calorimeter,
Heat lost = heat gained.
Let the final temperature of the mixture = 100ºC.
Heat gained by ice and iced water Q
1
is
Q
1
=m
i
L
i
+m
i
sΔT=200×80+200×1×100=3600cal.
Q
2
=(m
w
s+w
e
)×(100−0)=(250×1+50)×100=30000cal.
Q=Q
1
+Q
2
=36000+30000=66000cal.
If entire steam condensed the total heat given by it is
Q
3
=m
s
×L=200×540=108000cal
Now the total heat available is 108000 cal, but only 66000 cal are required. Therefore all the steam will not get condensed.
Final temperature of the mixture is 100ºC.
Mass of steam condensed is
540
66000
=122.2g.
Mass of the contents is 250+200+122.2=572.2g
Explanation:
a liquid of mass 240 gram contained in a calorie metre of water equivalent 12 gram at temperature for 14 degree Celsius A copper piece of mass 72 gram and specific heat 0.093 increased to 100 degree Celsius and then put into calorie metre if final temperature 18 degree Celsius find specific heat of liquid.
