Chapter 1 Physical Environment Solved Numerical Problems
Problem 1 How much heat will be required to raise the temperature of 200 grams of water from 30°C to 90°C? (Given: specific heat of water is 1 cal/g.°C)
Solution:
Given:
Temperature of 200 grams of water from 30°C to 90°C
Initial temperature of the water, t = 30°C
The final temperature of the water, t2 = 90°C
Change in temperature,
(t2-t1) = (90-30)°C = 60°C
Mass of water, m = 200 g
Specific heat of water, s = 1 cal/g.°C
The heat required = m.s. (t2 – t1) = 200 x 1 x 60 cal = 12000 cal.
Problem 2. A brass utensil of 400 g weight at 100°C cools down to 25°C by radiating heat. Calculate the amount of heat released by the brass utensil. (Given: specific heat of brass = 0.09 cal/g.°C)
Solution :
Given:
A brass utensil of 400 g weight at 100°C cools down to 25°C by radiating heat.
The initial temperature of brass, t1 = 100°C
The final temperature of brass, t2 = 25°C
Change in temperature, (t1-t2) = ( 100-25)°C = 75°C
Mass of brass, m = 400 g Specific heat of brass,
s = 0.09 cal/g.°C Heat rejected by the brass utensil = m.s.(t1 -t2)
= 400 x 0.09 x 75 cal = 2700 cal
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Problem 3. 100 grams of water is taken at 20°C temperature. If 50 grams of a substance at 100°C is dropped in that water, the final temperature becomes 50°C. Calculate the specific heat of that substance.
Solution:
Given:
100 grams of water is taken at 20°C temperature. If 50 grams of a substance at 100°C is dropped in that water, the final temperature becomes 50°C.
Heat gained by water,
= 100 x l x (50 – 20) cal = 3000 cal Heat released by the substance,
H2 = 50 x s x (100 – 50) cal,
where “s” is the specific heat of the substance
We know that, heat rejected = heat gained or, H2 = H1
or, 50 x s x (100 – 50) = 3000 cal
or,s =3000/50×50 =1.2 cal/g°.C
Chapter 1 Physical Environment Solved Numerical Problems
Problem 1. The latent heat for melting ice at 0°C and 1 atm is 80 cal/g. How much heat has to be supplied to completely convert 200g ice at 0°C to water at 0°C
Solution:
Given:
Latent heat for melting of ice at 0°C and 1 atm = 80 cal/g
The heat required to completely convert 1 g ice at 0°C to water at 0°C = 80 cal
So, the heat is required to completely convert 200 g ice at 0°C to water at 0°C = (80 x 200) cal = 16000 cal.
Problem 2. The latent heat for melting ice at 0°C and 1 atm is 80 cal/g. Calculate its value in the SI units.
Solution:
Given:
Latent heat for melting of ice at 0°C and 1 atm is 80 cal/g
The heat required to completely convert 1 g ice at 0°C to water at 0°C = 80 cal
So, the heat required to completely convert
1000 g ice at 0°C to water at 0°C
= (80 x 1000) cal = 80000 cal
We know that 1 cal = 4.2 J
Hence, 80000 cal
= (80000 x 4.2) J = 336000 J So, latent heat of melting of ice = 336000 J/kg