TG EAMCET 2 May 2025 Shift 1 Paper

Heat of raise temperature of ice from −20∘C to 0∘C.
Q1‌=m⋅C‌ice ‌⋅∆T
‌=0.008×2100×20=336J
Heat to melt ice at 0∘C to water
Q2=m×Lf=0.008×336000=2,688J
Heat to raise temperature of water from 0∘C to 100∘C
Q3‌=m×C‌water ‌‌‌×∆T
‌=0.008×4200×100=3360J
Heat to convert water at 100∘C to steam
Q4‌=M⋅Lv=0.008×2.268×106
‌=18144J
Total heat required
Q‌Total ‌‌=Q1+Q2+Q3+Q4
‌=336+2688+3360+18144
‌=24,528J(‌ or ‌24.5kJ)
1 calorie =4.18J
Total heat in joules
Q=‌

24528

4.2

≈5840cal