Question

Heat is added to 2.09 kg of ice at -8 °C. How many kilocalories are required to change the ice to steam at 131 °C?

Answer 1

the ice at -8 is taken heated to ice at 0 degrees, ice at 0 degres is heated to water at 0degrees, water at 0 degrees is heated to water at 100 , water at 100 is heated to steam at 100 , steam at 100 is heated to steam at 131

sice = 0.5 kcal/lg

Sw = 1 kcal / kg

Lice = Lf = 80 kcal/Kg

Lwater = Lv = 540 kcal/kg

Ssteam = 0.48 kcal/kg

heat reuired to take ice at -8 to ice at 0 = Q1 = m**sice*dT = 2.09*0.5*8 = 8.36 kcal

heat required to take ice at 0 to water at 0 = Q2 = m*Lf = 2.09*80 = 167.2 Kcal

heat required to take water at 0 to water at 100 = Q3 = m**sw*dT = 2.09*1*100 = 209

heat required to take water at 100 to steam at 100 = Q4 = m**Lv = 2.09*540 = 1128.6 Kcal

heat reuired to take steam at 100 to steam at 131 = Q5 = m**Ssteam*dT = 2.09*0.48*31 = 31.1 Kcal

totqal heat to be added = Q + Q2 +Q3 + Q4 +Q5

Q = 8.36+167.2+209+1128.6+31.1

Q = 1544.26 Kcal   <-------answer