Question

Calculate the amount of heat required to convert 1.00 kg of ice at -10°C at normal pressure to steam at 100°C. Specific heat capacity of ice = 2100 J kg⁻¹ K⁻¹ , latent heat of fusion of ice = 3.36 × 10⁵ J kg⁻¹ , specific heat capacity of water = 4200 J kg ⁻¹ K⁻¹ and latent heat of vaporization of water = 2.25 × 10⁶ J kg⁻¹

• A. $3.03\times {10}^{6}J$
• B. $3.03\times {10}^{12}J$
• C. $3.303\times {10}^{-6}J$
• D. $3.303\times {10}^{5}J$

Answer 1

The correct option is A $3.03\times {10}^{6}J$

Heat required to take the ice from -10°C to 0°C
= (1 kg) (2100 J kg⁻¹ K⁻¹) (10K) = 21000 J.
Heat required to melt the ice at 0°C to water
= (1 kg) (3.36 × 10⁵ J kg⁻¹) = 336000 J.
Heat required to take 1 kg of water from 0°C to 100°C
= (1 kg) (4200 J kg⁻¹ K⁻¹) (100 K) = 420000 J.
Heat required to convert 1 kg of water of 100°C into steam
= (1 kg) (2.25 × 10⁶ J kg⁻¹) = 2.25 × 10⁶J

Total heat required = 3.03 × 10⁶ J