Question

calculate heat energy required to change 12g of ice at - 10 ° C into water vapour
at 100 c​

Answer 1

9571680 J heat required to change 12g of ice at $-10^0C$ into water vapour at

$100^0$ C.

1) given information in question

  mass of ice = 12 g= 0.012 g

temperature of ice ΔT = -$10^0 c$

changing ice into water vapour temperature Δ$T_{1}$ = $100^0 C$

2) constant values

  $C_{w} =4186 J/Kg.K\\C_{w} = 2302J/Kg.K\\L_{f} =3.33 10^5 J/kg\\Heat energy Q =?\\\\ Heat enegy Q = Q_{1}+Q_{2}+Q_{3}\\Now \\                     Q_{1} =m.C_{i}.DT\\                     Q_{1} =12*2302*20 = 552480 \\                         Q_{2} = m.L_{f}                       Q_{2} = 12*3.33 10^5 =3996000\\\\                       Q_{3} =m*C_{w}*DT_{1}                        Q_{3} = 12*4186*100 = 5023200\\   total required energy Q = 552480+3996000+5023200$

                         Q = 9571680 J energy required to form water vapour of -10*C ice.