Q1: Convert 30°C into Kelvin and Fahrenheit Scale.
Given:
Temperature in Celsius = 30°C
Formulas:
Kelvin = °C + 273
Fahrenheit = (°C × 9/5) + 32
Calculation (Kelvin):
K = 30 + 273
K = 303 K
Calculation (Fahrenheit):
°F = (30 × 9/5) + 32
°F = (30 × 1.8) + 32
°F = 54 + 32
°F = 86°F
Answer: 303 K and 86°F
Temperature in Celsius = 30°C
Formulas:
Kelvin = °C + 273
Fahrenheit = (°C × 9/5) + 32
Calculation (Kelvin):
K = 30 + 273
K = 303 K
Calculation (Fahrenheit):
°F = (30 × 9/5) + 32
°F = (30 × 1.8) + 32
°F = 54 + 32
°F = 86°F
Answer: 303 K and 86°F
Q2: Convert 212°F into Celsius and Kelvin.
Given:
Temperature in Fahrenheit = 212°F
Formulas:
Celsius = (°F - 32) × 5/9
Kelvin = °C + 273
Calculation (Celsius):
°C = (212 - 32) × 5/9
°C = 180 × 5/9
°C = 180 × 0.5556
°C = 100°C
Calculation (Kelvin):
K = 100 + 273
K = 373 K
Answer: 100°C and 373 K
Temperature in Fahrenheit = 212°F
Formulas:
Celsius = (°F - 32) × 5/9
Kelvin = °C + 273
Calculation (Celsius):
°C = (212 - 32) × 5/9
°C = 180 × 5/9
°C = 180 × 0.5556
°C = 100°C
Calculation (Kelvin):
K = 100 + 273
K = 373 K
Answer: 100°C and 373 K
Q3: How much heat is required to boil 3 kg water which is initially 10°C?
Given:
Mass of water, m = 3 kg
Initial temperature, T₁ = 10°C
Final temperature (boiling), T₂ = 100°C
Specific heat capacity of water, c = 4200 J/kg°C
Formula: Q = m × c × ΔT
where ΔT = T₂ - T₁
Calculation:
ΔT = 100 - 10 = 90°C
Q = 3 × 4200 × 90
Q = 3 × 4200 × 90
Q = 3 × 378000
Q = 1,134,000 J
Q = 1.134 × 10⁶ J
Answer: 1.134 × 10⁶ Joules
Mass of water, m = 3 kg
Initial temperature, T₁ = 10°C
Final temperature (boiling), T₂ = 100°C
Specific heat capacity of water, c = 4200 J/kg°C
Formula: Q = m × c × ΔT
where ΔT = T₂ - T₁
Calculation:
ΔT = 100 - 10 = 90°C
Q = 3 × 4200 × 90
Q = 3 × 4200 × 90
Q = 3 × 378000
Q = 1,134,000 J
Q = 1.134 × 10⁶ J
Answer: 1.134 × 10⁶ Joules
Q4: 2kg of copper requires 2050J of heat to raise its temperature through 10°C. Calculate the heat capacity of the sample.
Given:
Mass of copper, m = 2 kg
Heat supplied, Q = 2050 J
Change in temperature, ΔT = 10°C
Formula: Heat Capacity = Q / ΔT
Calculation:
Heat Capacity = 2050 / 10
Heat Capacity = 205 J/°C
Answer: 205 J/°C
Mass of copper, m = 2 kg
Heat supplied, Q = 2050 J
Change in temperature, ΔT = 10°C
Formula: Heat Capacity = Q / ΔT
Calculation:
Heat Capacity = 2050 / 10
Heat Capacity = 205 J/°C
Answer: 205 J/°C
Q5: An iron block of volume 3m³ is heated, so that its temperature changes from 25°C to 100°C. If the coefficient of linear expansion of iron is 11 × 10⁻⁶ °C⁻¹. What will be the new volume of the iron block after heating?
Given:
Initial Volume, V₀ = 3 m³
Initial temperature, T₁ = 25°C
Final temperature, T₂ = 100°C
Coefficient of linear expansion, α = 11 × 10⁻⁶ °C⁻¹
Formulas:
β (coefficient of volume expansion) = 3α
ΔT = T₂ - T₁
V = V₀(1 + βΔT)
Calculation:
β = 3 × (11 × 10⁻⁶)
β = 33 × 10⁻⁶ °C⁻¹
β = 3.3 × 10⁻⁵ °C⁻¹
ΔT = 100 - 25 = 75°C
V = 3 [1 + (3.3 × 10⁻⁵ × 75)]
V = 3 [1 + (3.3 × 10⁻⁵ × 75)]
V = 3 [1 + (247.5 × 10⁻⁵)]
V = 3 [1 + 0.002475]
V = 3 × 1.002475
V = 3.007425 m³
Answer: 3.0074 m³
Initial Volume, V₀ = 3 m³
Initial temperature, T₁ = 25°C
Final temperature, T₂ = 100°C
Coefficient of linear expansion, α = 11 × 10⁻⁶ °C⁻¹
Formulas:
β (coefficient of volume expansion) = 3α
ΔT = T₂ - T₁
V = V₀(1 + βΔT)
Calculation:
β = 3 × (11 × 10⁻⁶)
β = 33 × 10⁻⁶ °C⁻¹
β = 3.3 × 10⁻⁵ °C⁻¹
ΔT = 100 - 25 = 75°C
V = 3 [1 + (3.3 × 10⁻⁵ × 75)]
V = 3 [1 + (3.3 × 10⁻⁵ × 75)]
V = 3 [1 + (247.5 × 10⁻⁵)]
V = 3 [1 + 0.002475]
V = 3 × 1.002475
V = 3.007425 m³
Answer: 3.0074 m³
