StudentAidUnit 10: Thermal Physics — Numericals
10th Class Physics · Unit 10: Thermal Physics
10.1.A metal rod of length 1 m expands by 0.02 m when heated from 20 °C to 120 °C. Calculate its coefficient of linear expansion.
Given
Initial length of rod
L0 = 1.00 m
Increase in length
ΔL = 0.02 m
Initial temperature
T0 = 20^circ C
Final temperature
T = 120^circ C
Temperature change
ΔT = T - T0 = 120^circ C - 20^circ C = 100^circ C or 100 K
To find
Coefficient of linear expansion
α = ?
Formula
Linear thermal expansion formula
α = ΔLL0 × ΔT
Working
Substituting values
α = 0.021.00 × 100
Simplification
α = 0.02100
Result
Final result
α = 2.0 × 10-4 K-1
10.2.A container holds 1 litre of water at 20 °C. What will be its volume at 80 °C, assuming water's coefficient of volume expansion is 2.1×10⁻⁴ per °C?
Given
Initial volume of water
V0 = 1.00 L = 1000 cm3
Coefficient of volume expansion
β = 2.1 × 10-4 /^circ C
Initial temperature
T0 = 20^circ C
Final temperature
T = 80^circ C
Temperature change
ΔT = T - T0 = 80^circ C - 20^circ C = 60^circ C
To find
Final volume
V = ?
Formula
Volume expansion formulas
ΔV = β × V0 × ΔT; quad V = ΔV + V0
Working
Substituting values for volume change
ΔV = 2.1 × 10-4 × 1000 × 60
Calculating volume change
ΔV = 2.1 × 10-4 × 60000 = 12.6 cm3
Calculating final volume
V = 12.6 + 1000 = 1012.6 cm3
Result
Final result
V = 1012.6 cm3
Alternative method
The problem can also be solved by V = V0(1 + β ΔT)
10.3.A steel rod initially measures 2 m at 20 °C. If its coefficient of linear expansion is 1.2 × 10⁻⁵ °C⁻¹, what will be its length at 100 °C?
Given
Initial length of rod
L0 = 2.00 m
Coefficient of linear expansion
α = 1.2 × 10-5 K-1
Initial temperature
T0 = 20^circ C
Final temperature
T = 100^circ C
Temperature change
ΔT = T - T0 = 100^circ C - 20^circ C = 80^circ C
To find
Final length
L = ?
Formula
Linear expansion formula
L = L0 × (1 + α × ΔT)
Working
Substituting values
L = 2.00 × (1 + 1.2 × 10-5 × 80)
Calculating the product in brackets
1.2 × 10-5 × 80 = 96 × 10-5 = 0.00096
Adding 1
1 + 0.00096 = 1.00096
Multiplying by initial length
L = 2.00 × 1.00096 = 2.00192 m
Result
Final result
L = 2.00192 m
10.4.A steel bridge expands by 5 cm on a hot summer day. If the bridge originally spanned 100 m, what is the temperature change?
Given
Original length of bridge
L0 = 100.0 m
Expansion in length
ΔL = 5 cm = 0.050 m
Coefficient of linear expansion
α = 1.2 × 10-5 K-1
To find
Temperature change
ΔT = ?
Formula
Linear expansion formula rearranged
ΔL = α × L0 × ΔT; quad or quad ΔT = frac{ΔL}{α × L0
Working
Substituting values
ΔT = 0.0501.2 × 10-5 × 100.0
Calculating denominator
1.2 × 10-5 × 100.0 = 1.2 × 10-3 = 0.0012
Division
ΔT = 0.0500.0012 = 41.67^circ C
Result
Final result
ΔT = 41.67^circ C
10.5.How much heat is required to raise the temperature of a 2 kg iron bar from 20 °C to 100 °C, given that the specific heat capacity of iron is 450 J kg⁻¹ K⁻¹?
Given
Mass of iron bar
m = 2.00 kg
Specific heat capacity of iron
c = 450 J kg-1 K-1
Initial temperature
T0 = 20^circ C
Final temperature
T = 100^circ C
Temperature change
ΔT = T - T0 = 100^circ C - 20^circ C = 80^circ C or 80 K
To find
Heat required
Q = ?
Formula
Heat formula
Q = m × c × ΔT
Working
Substituting values
Q = 2 × 450 × 80
Calculation
Q = 900 × 80 = 72000 J
Converting to kilojoules
Q = 720001000 = 72 kJ
Result
Final result
Q = 72 kJ
10.6.How much heat is required to melt 500 g of ice at 0°C into water at 0°C? (Latent heat of fusion of ice = 3.36 × 10⁵ J kg⁻¹ = 0.5kg)
Given
Mass of ice
m = 500 g = 5001000 = 0.5 kg
Latent heat of fusion of ice
Lf = 3.36 × 105 J kg-1
To find
Heat required to melt ice
Q = ?
Formula
Latent heat formula
Q = m × Lf
Working
Substituting values
Q = 0.5 × 3.36 × 105
Calculation
Q = 0.5 × 336000 = 168000 J
Converting to kilojoules
Q = 168 kJ
Result
Final result
Q = 168 kJ
10.7.Calculate the heat required to completely vaporize 1 kg of water at 100 °C. (Latent heat of vaporization of water = 2.26 × 10⁶ J kg⁻¹)
Given
Mass of water
m = 1.00 kg
Latent heat of vaporization of water
Lv = 2.26 × 106 J kg-1
To find
Heat required to vaporize water
Q = ?
Formula
Latent heat formula
Q = m × Lv
Working
Substituting values
Q = 1 × 2.26 × 106
Calculation in joules
Q = 2.26 × 106 J
Converting to megajoules
Q = 2.26 MJ
Result
Final result
Q = 2.26 × 106 J = 2.26 MJ