Answer:
Amount of heat required = 54601.2 J
Amount of heat required = 13050 cal
Amount of heat required = 51.68 Btu
Explanation:
Mass of ice, m = 18.0 g
Initial temperature of ice, T₀ = -10.0 ⁰C
Specific heat of ice, C₀ = 0.50 cal/g-°C
Final temperature of ice, T₁ = 0 ⁰C
Amount of heat required to change the temperature of ice from T₀ to T₁ is:
Q₁ = mC₀( T₁ - T₀)
Q₁ = 18 x 0.50 x ( 0 + 10 )
Q₁ = 90 cal
Latent heat of ice, L₁ = 80 cal/g
Amount of heat required to change ice into water at T₁ temperature is:
Q₂ = m x L₁
Q₂ = 18 x 80 = 1440 cal
Final temperature of water, T₂ = 100 °C
Specific heat of water, C₁ = 1 cal/g-°C
Amount of heat required to change the temperature of water from 0 °C to 100 °C, that is, from T₁ to T₂ is:
Q₃ = mC₁(T₂ - T₁)
Q₃ = 18 x 1 x (100 - 0)
Q₃ = 1800 cal
Latent heat for boiling, L₂ = 540 cal/g
Amount of heat required to change water into steam at 100 °C is:
Q₄ = mL₂
Q₄ = 18 x 540 = 9720 cal
Total amount of heat required to change ice at -10 °C to steam at 100 °C is:
Q = Q₁ + Q₂ + Q₃ + Q₄
Q = 90 + 1440 + 1800 + 9720
Q = 13050 cal
But, 1 cal = 4.184 joule
So, in joules the heat required is:
Q = 13050 x 4.184 = 54601.2 J
1 cal = 3.96 x 10⁻³ Btu
In terms of Btu, the heat required is:
Q = 13050 x 3.96 x 10⁻³ = 51.68 Btu
