Answer:
a) Q = 80,000 cal
b) Q = 100,000 cal
c) Q = 540,000 cal
d) Q = 720,000 cal
Explanation:
a)1 kg from 0⁰ Ice to 0⁰ water, the heat produced is latent heat of fusion
[tex]Q_{l} = ML_{f}[/tex] = 1 * 80
[tex]Q_{l}[/tex] = 80 kCal = 80,000 cal
b) 1 kg of O°C ice water to 1 kg of 100°C boiling water
Specific heat capacity, c = 1000cal/kg.C
[tex]Q_{c} = mc \delta T\\Q_{c} = 1 * 1000 * (100 - 0)\\Q_{c} =100000 cal[/tex]
c) 1 kg of 100°C boiling water to 1 kg of 100°C steam
Latent heat of vaporization is needed for this conversion
[tex]Q_{v} = ML_{v} \\L_{v} = 540 kCal/kg\\Q_{v} =1* 540 \\Q_{v} = 540 kCal = 540000 cal[/tex]
d) 1 kg of O°C ice to 1 kg of 100°C steam.
Q = [tex]Q_{L} + Q_{c} + Q_{v}[/tex]
Q = 80,000 + 100,000 + 540,000
Q = 720,000 cal