CBSE Class 12 Physics Calorimetry Solved Examples

CBSE Class 12 Physics Calorimetry Solved Examples. Please refer to the examination notes which you can use for preparing and revising for exams. These notes will help you to revise the concepts quickly and get good marks.

Ex1. 1 g of steam at 100°C can melt how much ice at 0°C? Latent heat of ice = 80 cal/g and latent heatof steam = 540 cal/g.
Sol. Heat required by ice for melting of m g of ice =
mL = m × 80 cal
Heat available with steam for being condensed and then brought to 0°C
= 1 × 540 × 100
= 640 cal
m × 80 = 640
or m = 640/80 = 8 grams

Ex 2. A tap supplies water at 10°C and another tap at100°C. How much hot water must be taken sothat we get 20 kg of water at 35°C ?
Sol. Let mass of hot water = m kg
mass of cold water
= (20 – m) kg
Heat taken by cold water
= (20 – m) × 1 × (35 – 10)
Heat given by hot water
= m × 1 × (100 –35)
Law of mixture givesHeat given by hot water
= Heat taken by cold water
m × 1 × (100 – 35) = (20 – m) × (35 – 10)
65 m = (20 – m) × 25
65 m = 500 – 25 m
or 90 m = 500
m =500 /90 = 5.56 kg

Ex 3. 5 g of ice at 0°C is dropped in a beaker containing 20 g of water at 40°C. What will be the finaltemperature?
Sol. Let final temperatue be = θ
Heat taken by ice = m1L + m1c1 Δ θ 1
= 5 × 80 + 5 × 1 ( θ – 0)
= 400 + 5 θ
Heat given by water at 40°C
= m2c2 Δ θ 2 = 20 × 1 × (40 – θ )
= 800 – 20 θ
As Heat given = Heat taken
800 – 20 θ = 400 + 5 θ
20 θ = 400
θ = 400/25

= 16° C

Ex 4. 5 g ice of 0°C is mixed with 5 g of steam at 100°C. What is the final temperature?
Sol. Heat required by ice to raise its temperature to 100°C,
Q1 = m1L1 + m1c1 Δ θ 1
= 5 × 80 + 5 × 1 × 100
= 400 + 500 = 900 cal
Heat given by steam when condensed,
Q2 = m2L2
= 5 × 536 = 2680 cal
As Q2 > Q1. This means that whole steam is not even condensed.
Hense temperature of mixture will remain at 100°C