Question

Using the information in Table 6.8 in the "Additional items" content section in D2L, calculate the total amount of heat needed to change 500 g of ice at -10°C into 500 g of steam at 120°C. Do this by calculating the heat required for each of the following steps and adding to get the total:

Step 1. Ice (-10°C) → ice (0°C)

Step 2. Ice (0°C) → water (0°C)

Step 3. Water (0°C) → water (100°C)

Step 4. Water (100°C) → steam (100°C)

Step 5. Steam (100°C) → steam (120°C)

a) 367 kcal

b) 417 cal

c) 103 cal

d) 54 kcal

Answer 1

Step-1 (Ire (-10%) - ice coe). Specific heat of ice a o.seal lare So total heat required for soog of ice is TH, = (500 xos x 10 Deal OH, = 3500 cal a 25 kead Step-2 (ree (0%) - water (ope)). lextent of a ice 80 caly- so OH, = (80x500) 40000 cal noked Step-3 (water (oc) -) coater (100c)). specific heat of water a lead Igre AH 2 8 (soo x1000)) cal- - soooo cal - sokeal Step-y (water (100'e) —) Steam (1000)) Latent heat of water to turn to Steama scocally 4 Hg = (500x suo) cal = 270000 cal Step-s ( Steam (1008) -) Steam (120°)) ²2 specific heat of steam= 0.48 cal/ge so Alts= (500 x 0,48 x 20) eal = 4.8 keal! So Total bento (94, +4,+ oHy+04, + avg) = 367.3 kcal/19)acial 2 270 Keal