Question

Heat is added to 1.79 kg of ice at -14 °C. How many kilocalories are required to change the ice to steam at 127 °C?

Answer 1

Q1= heat required to raise temperature of ice upto 0 degree

= mC(t2-t1) = 1.79×2108×(0-(-14))= 52826.6 J

Q2= heat required to melt ice at 0 degree =mL_f

=1.79×334000= 597860 J

Q3= heat required to heat water from 0 degree tp 100 degree

= ms(t3-t2) = 1.79×4180×100= 748220 J

Q4= heat required to vapourise water at 100 degree

= mL_v = 1.79×2230000= 3991700J

Q5= heat required to heat steem from 100 degree to 127 degree = ms(t4-t3) = 1.79× 1996× 27= 96466.7 J

Total heat required is Q= Q1+Q2+Q3+Q4+Q5

Q= 5487073.3J = 5487083.3/4183= 1311.8 kilo calories

Q= 1311.8 kilo calories