calcilate the quantity of heat required to convert 2kg of ice at -20degree celcius changes into 120degree celcius steam ?
Answers
Answer:
Heat energy required to convert ice at –10°C to ice at 0°C is given by:
Q = msΔβ.
Q = 0.02 × 2090 × (0 – (–10)).
Q = 418J.
Heat energy required to convert ice at 0°C to water at 0°C is given by:
Q = ml.
l = specific latent heat of fusion of ice = 336000Jkg-¹.
Q = 0.02 × 336000.
Q = 6720J.
Heat energy required to convert water at 0°C to water at 100°C is given by:
Q = msΔβ.
Q = 0.02 × 4200 × 100.
Q = 8400J.
Heat energy required to convert water at 100°C to steam at 100°C is given by:
Q = mL.
L = specific latent heat of vaporization of steam = 2260000Jkg-¹
Q = 0.02 × 2260000.
Q = 45,200J.
Heat energy required to convert steam at 100°C to steam at 120°C is given by:
Q = msΔβ.
s = specific heat capacity of steam = 1,996Jkg-¹K-¹.
Q = 0.02 × 1,996 × (120 – 100).
Q = 0.02 × 1,996 × 20.
Q = 798.4J.
Total heat energy supplied = 418J + 6720J + 8400J + 45200J + 798.4J = 61536.4J / 61.536kJ.