5) Show that Q total in cal is needed to change 50 g of 0 ̊C ice to steam at 100 ̊C.
(a) Show that Q in cal, Quantity of heat is needed to increase the temperature of 50 g water from 0 ̊C to 100 ̊C. The specific heat capacity for water is 1 cal/g•̊C.
(b) Show that Q in cal, Heat of fusion is needed to melt 50 g of 0 ̊C ice. The heat of fusion Lffor water is 80 cal/g.
(c) Show that Q in cal, Heat of vaporization is needed to turn 50 g of 100 ̊C boiling water into steam. The heat of vaporization for water Lv is 540 cal/g.
(d) Sum the heat of fusion, quantity of heat, and heat of vaporization in order to calculate the Q total in cal is needed.
A total of 581 cal of heat is added to 5.00g ice at -20 C. What is the final temperature of the ice? Specific heat of H2O (SOLID) = 2.087 Specific heat of H2O (Liquid)= 4.184 Heat of fusion of H2O = 333.6J/g Hint: The total amount of energy needed is equal to the sum of heat needed to warm the ice to 0.0 C, melt the ice and warm the water to its final temperature
8. Use the data in the Introduction calculate the total amount of heat in kcal required to turn 100 g of ice at -20°C to steam at 120°C? liq gas equilibrium (heat goes into phase change) Steam - Water and steam allas (heat goes into temperature change) Temperature (°C) all liquid (heat goes into temperature change) -Water Ice and water all solid Nice solid/liq equilibrium (heat goes into phase change) - Time Heat On the 5 sections of the graph...
The heat of vaporization is 540 cal/g. How many kilocalories are needed to change 5.2 g of liquid water to steam at 100°C. Treat as exact
The heat of vaporization of water is 540 cal/g, and the heat of fusion is 80 cal/g. The heat capacity of liquid water is 1 cal g−1 °C−1, and the heat capacity of ice is 0.5 cal g−1 °C−1. 18 g of ice at -6°C is heated until it becomes liquid water at 40°C. How much heat was required for this to occur?
Given that the specific heat capacities of ice and steam are 2.06 J/g°C and 2.03 J/g°C, the molar heats of fusion and vaporization for water are 6.02 kJ/mol and 40.6 kJ/mol, respectively, and the specific heat capacity of water is 4.18 J/g°C, calculate the total quantity of heat evolved when 24.1 g of steam at 158°C is condensed, cooled, and frozen to ice at -50.°C.
Calculate how many grams of ice at 0°C would be melted by 100 g of 100°C steam. Hint: heat will be transferred from the steam to the ice in two processes: •the steam will condense into liquid, and •that liquid will transfer heat until it is at the freezing point of water. Note that for water: Lf = 80 cal/g (ice), and Lv = 540 cal/g (steam) (Use calories) C(water) =1 cal/g*C Answer in grams no decimal places.
18 grams of ice at –31°C is to be changed to steam at 199°C. The entire process requires _____ cal. Round your answer to the nearest whole number. The specific heat of both ice and steam is 0.5 cal/g°C. The specific heat of water is 1.00 cal/gK. The heat of fusion is 80 cal/g and the heat of vaporization is 540 cal/g.
Q13) 17 grams of ice at –35°C is to be changed to steam at 179°C. The entire process requires _____ cal. Round your answer to the nearest whole number. The specific heat of both ice and steam is 0.5 cal/g°C. The specific heat of water is 1.00 cal/gK. The heat of fusion is 80 cal/g and the heat of vaporization is 540 cal/g.
9) How much energy in Kcal is needed to take 340 g of water at 6.0 degrees Celsius to steam at 127.0 degrees Celsius? ALSO indicate this process as a heating curve on the axes below. [the specific heat of ice, water and steam are 0.47, 1.00, and 0.48 cal/g oC respectively, Also the Heat of Fusion and Heat of vaporization of water are 80, and 540 cal/g respectively] You may or may not need all of these numbers. hny...
How many calories are needed to convert 125 grams of water at 75.0 C to steam at 100.0 C? [Specific heats: ice = 0.495, water = 1.00 steam = 0.478 cal/g C] [water's heat of fusion = 80.0 cal/gram, water's heat of vaporization - 540.0 cal/gram] 3125 cal 3330 cal 6875 cal 13,125 cal 70,625 cal