Question

Question 20 (8 points) The energy required to convert 18.0 g ice at -25.0 C to water at 50.0C is _ _kl. Specific heat capacity of ice (2.03 J/g °C); AHfus of water (6.01 kJ/mol) Specific heat capacity of liquid water (4.18 J/g °C); AHvap of water (40.67 kJ/mol) Specific heat of steam 1.84 J/g °C (Hint: you will not be using all the data given here.) 8.83 O 47.1 12.3 O 10.7 06.27

Answer 1

Solution is :

2) Useful Information: Specific heat of ice = 2.03 Ilgic Specific heat of water = 4.18 Ilgie Heat of fusion of water = 334 Fly The total energy required is the sum of the energy to heat the – 25 c ice to oc ice, melting the o'c into oc water heating the water to 50°c. Step 1 Heat required to raise the temperature of ice from – 25°C to 0°c Since, q = MCAT where qa heat energy m= mass a specific heat AT = Change in Temperature. By substituting Values q = (18g) x (2.03 J/g °C) x [0°c - (-25°c)] 2 = (189) * (2.03 T /g °C) x 25°C 9 = 913.5 T Step 2: Heat Required to convert oc ice to oc water Since, q = m. Alf where qa heat energy m = mass Alf - heat of fusion 9 = (189 x 334 Jlg ) = 6012 T

Step 3: of oc water Heat required to raise the temperature to 50 c water = meat 9 = (189 )* (4.18 T/s°C) [50*2-0°C] 22 (189 x 4.18 J/g °c x 50°C q= 3762 I eneran will be = - 913.5+ 6012 T + 3762 I 1068 7.5 T = = 10.6875KT 10.7 km So, option d) is correct.