1:13 0* ll 66%iCyra GarcesActive NowCALORIMETRYHEAT OF FUSIONTrial 1Trial 2Mass of ice M (grams)7979Mass of warm waterMter (grams)120119Initial temperature of waterTinitial ("C)4142Final temperature of mixtureTfinal ("C)-4Calculated heat of fusion of iceLeperinemtal (cal/g)72.460.8Standard value of heat offusion of ice Lemdard(cal/g)79.779.7Percentage error9.22%23.76%CalculationsAccording to the principle of the conservation of energy, the quantity of heat absorbed by the ice asit melts (Quained melring) and then heats up to the final equilibrium temperature (Qpain by meltedice) must equal the quantity of heat released by the warm water (Qlast by warm water) as it cools downto the final equilibrium temperature.Mathematically:leguined metingl iast by warm waterpained meltingl + ocat(M(L) + (M)(1)T - OC) = (Mt(1)(Tie- Tfina- 0C) - (Mt(1finalwaterinitialMoreEditII- 3. Make the step by stepcomputation for trial 2,

In calorimetry, the heat lost by one substance is gained by the other substance as there is no loss…

Trial 1 Trial 2 Mass of ice M.. (grams) 79 79 Mass of warm water 120 119 Mer (grams) Initial temperature of water Tinitial ("C) 41 42 Final temperature of mixture Tfinal ("C) 4 Calculated heat of fusion of ice 72.4 60.8 amerinental (cal/g) Standard value of heat of fusion of ice Lutmdard (cal/g) 79.7 79.7 Percentage error 9.22% 23.76%

Trial 1 Trial 2 Mass of ice M.. (grams) 79 79 Mass of warm water 120 119 Mer (grams) Initial temperature of water Tinitial ("C) 41 42 Final temperature of mixture Tfinal ("C) 4 Calculated heat of fusion of ice 72.4 60.8 amerinental (cal/g) Standard value of heat of fusion of ice Lutmdard (cal/g) 79.7 79.7 Percentage error 9.22% 23.76%

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Similar questions

Equations The heat energy associated with a change in temperature that does not involve a change in phase is given by q=msAT where q is heat in joules, m is mass in grams, s is specific heat in joules per gram-degree Celsius, J/(g · °C), and AT is the temperature change in degrees Celsius. The heat energy associated with a change in phase at constant temperature is given by q=mΔΗ where q is heat in joules, m is mass in grams, and AH is the enthalpy in joules per gram. Physical constants The constants for H₂O are shown here: ● ● Specific heat of ice: Sice = 2.09 J/(g. °C) Specific heat of liquid water: swater = 4.18 J/(g · °C) Enthalpy of fusion (H₂O(s)→H₂O(1)): AHfus = 334 J/g Enthalpy of vaporization (H₂O(1)→H₂O(g)): AHvap = 2250 J/g
There are 270 Calories in a Hershey's® Milk Chocolate bar. The following data might be useful: AHvap = 43.9 kJ/mol, AHfusion = 6.02 kJ/mol, AHsublimation = 49.9 kJ/mol, specific heat of ice = 2.05 J g ¹°C 1, specific heat of water = 4.18 Jg ¹1°C-1, and specific heat of steam = 2.08 Jg ¹¹°C 1. Assume the specific heat of water does not change with temperature. (Density of ice = 0.920 g/mL). X Incorrect. How much energy is used up at each step of the heating, including heating the solid, melting, heating the liquid, and vaporization? If all of the energy in the bar were used to heat up 75.0 mL of ice at -25.0 °C, how hot would the water get? i 6406 ! °C
None
Answer “C”
q.6 How much energy (J) is needed to melt a 7.0 lb bag of ice? q.7 Calculate the energy needed to melt 3.0 g of ice at 0°C, heat it to boiling, boil it, and heat the stear to 125°C? 1
If 100 grams of ice at -10.0 degrees C is placed in 200 grams of water at 90 degrees C in an insulated container, what will be the temperature of the system when equilibrium is established? Specific heat of solid H2O = 2.09 J/g*deg C Specific heat of liquid H2O = 4.18 J/g*deg C Heat of fusion of H2O soliid = 333 J/g -- Another one I've worked several times but cannot come up with a reasonable solution.
A 0.517 g sample of steam at 104.0 °C is condensed into a container with 5.53 g of water at 14.5 *C. What is the final the specific heat of steam is temperature of the water mixture if no heat is lost? The specific heat of water is 4.18 2.01 and AHvap = 40.7 kJ/mol. T₁ = °℃
What is the enthalpy change for converting 10.0 g of ice at -15.0 °C to water at 55.0 °C in kJ. The specific heats of ice and water are 2.09 J/g-K and 4.18 J/g-K, respectively. For H20, AHfus = 6.01 kJ/mol, and AHvap = 40.67 kJ/mol %3D (note: your answer must be in kJ)
Determine the total amount of heat in kJ needed to warm 29 g of ice at -20.9 °C to water at 38.3 °C. Specfic heat capacities: Cice = 2.108 J/g °C Cwater = 4.184 J/g °C heat of fusion = 335 J/g Freezing point of water is 0 °C
2:42 PM Mon Jan 30 The normal boiling point of a liquid is 282.0 °C. What is the vapor pressure of this liquid at 136.0 °C in atm? (AHvap = 28.5 kJ/mol) Tap here or pull up for additional resources Question 18 of 20 @ (T)
Question 24 Shown below is the heating curve (q vs. T) for some substance. (3) D 20- heat added (kl/ml) melting point enthalpy of fusion -35. "C 5.00 kJ/mol 2.90 g/cm³ (solid) boiling point enthalpy of vaporization density 25. "C 30.00 kJ/mol 28. J-K mel (sold) 2.20 g/ml. (quid) heat capacity 56. JK mel 30. JK mel (liquid) (vapor) Which value was used to determine line segment AB? O 5.00 kJ/mol 30.00 kJ/mol 28 J K¹ mol¹ 30 J K¹¹ mol-1 56 J K¹ mol¹
How many grams of benzene, C,H,, can be melted with 88.8 kJ of heat energy? The molar heat of fusion of benzene is 9.95 kJ/mol. * TOOLS x10