Q.53 The heat required to convert 2 kg of water at 20 °C in a calorimeter to steam at 100
°C and at atmospheric pressure (1 atm) is _______ kJ. (Specific heat capacity of
water is 4.2 kJ kg–1 K–1 and latent heat of steam is 2256 kJ kg–1)
Step-by-Step Calculation
Convert water at 20°C to steam at 100°C in two phases: sensible heat to reach boiling point, then latent heat for phase change.
Heat for Temperature Rise
Q₁ = m × c × ΔT
Mass m = 2 kg, specific heat c = 4.2 kJ kg⁻¹ K⁻¹, ΔT = 100 – 20 = 80 K.
Q₁ = 2 × 4.2 × 80 = 672 kJ.
Heat for Vaporization
Q₂ = m × L
Latent heat L = 2256 kJ kg⁻¹ at 100°C and 1 atm.
Q₂ = 2 × 2256 = 4512 kJ.
Total Heat
The total heat required is 5184 kJ. This calculation accounts for heating 2 kg of water from 20°C to 100°C and then vaporizing it at 100°C under 1 atm pressure.
Q = Q₁ + Q₂ = 672 + 4512 = 5184 kJ.
Common Errors Explained
- No options provided, but typical mistakes include ignoring one phase or unit mismatches.
- Forgetting sensible heat: Yields only 4512 kJ (underestimates by 672 kJ).
- Wrong specific heat (e.g., 4.184 instead of 4.2): Minor difference (~4 kJ), but problem specifies 4.2.
- Using cal/g values: Latent heat ~540 cal/g = 2259 kJ/kg (close, but use given 2256).
- Superheating steam: Unneeded, as target is steam at 100°C.
Full Solution Breakdown
As computed: Sensible heat (672 kJ) + latent heat (4512 kJ) = 5184 kJ. Verify with formula Q = m c ΔT + m L
Practical Applications
- Boiler design and calorimetry experiments.
- CSIR NET problem-solving for phase changes.
This matches exam standards, avoiding pitfalls like unit conversion errors.
