Calculate the ideal power in kW to make 1 Ton (=1000 kg) of ice
at 0 deg.C per hour from water at 20 deg.C. Surroundings at 20
deg.C.
Use the expression for Wideal to calculate the power.
The total enthalpy change of the water = [(enthalpy change of the
water cooled from 20 to 0 deg.C) + (enthalpy change of water to ice
at 0 deg.C, this is the heat of fusion of water)].
The total entropy change of the water = [(entropy change of water
cooled from 20 to 0 deg.C) + (entropy change of ice at 0 deg.C =
(heat of fusion of water)/(temperature of fusion of water to ice)].
You can find enthalpy and entropy changes of the water from a steam
table. Be sure to keep the signs of the terms correct. The answer
is about 12 kW.
Calculate the ideal power in kW to make 1 Ton (=1000 kg) of ice at 0...
1. Ice (1.0 kg) at -10.0 °C is allowed to melt and warm to 25.0 °C, under isobaric conditions (1 bar). Considering the ice/water as the system, calculate the entropy changes for both the system and the surroundings. What is the overall change in entropy? Over the temperature ranges considered, the heat capacities of ice and water can be considered constant, at 36.9 J mol-1 K-1 for ice and 75.29 J mol-1 K-1 for water at 1 bar. The enthalpy...
You have a block of ice at a temperature of -100°C. This block of ice is made from 180g H2O. The block of ice will be heated continually until it becomes super-heated steam at a temperature of 200°C Cice = 2.03 J/g-K ΔHfus=6.01 kJ/mol Cwater = 4.18 J/g-K Csteam = 1.84 J/g-K ΔHvap=40.67 kJ/mol What is the enthalpy change raising the temperature of 180 g of ice at −100 °C to 0°C? What is the enthalpy change upon melting 180...
An ice tray contains 750g of liquid water at 0C. A) Calculate the entropy change of water as it is cooled slowly from 30C to 0C. B) Calculate the change in entropy of the water as it freezes slowly and completely at 0C. The heat of fusion of water is 3.33*10^5 j/kg
I would really appreciate your help. God bless you Question 4 The molar enthalpy of fusion of ice at 0 °C and 1 atm pressure is 6024 J mol. The molar heat capacities at constant pressure of ice and water are 37.65 J Kmol and 75.30 J Kmol respectively, and may be taken as constant over the temperature range 0 to -20°C. Consider 2 mole of liquid water supercooled to -20 °C, which is allowed to freeze isothermally from liquid...
.1. A 2.50 kg block of ice at 0°C is added to a picnic cooler. How much heat (in kcal) will the ice remove as it melts to water at 0°C? 2. A 300.0 g pot of water at room temperature (25.0°C) is placed on a stove. How much heat (in kcal) is required to change this water to steam at 100.0°C? 3. Spent steam from an electric generating plant leaves the turbines at 110.0°C and is cooled to 95.0°C liquid water...
Calculate the change in entropy when 55 g of ice at 0 ℃ melts to water at 0 ℃? The heat of fusion is 3.34x10 5 J/kg.
Calculate the change in entropy when 65 g of ice at 0 °C melts to water at 0 °C? The heat of fusion is 3.34x10 J/kg.
Please help 1. Calculate the increase of entropy (in J/K) when 42 g of ice melts at 0 ºC and 1 atm. (The heat of fusion for ice is 6,000 J/mol.) 2. Calculate the change in entropy (in J/K) when a 34.0 g of water is heated from 12.4 ºC to 70.5 ºC at 1 atm. (The specific heat is 4.184 J/(g-K).) Notice that entropy and heat capacity have the same units.
Calculate the entropy change when 72.00 g of ice, at 273.2 K and 1.000 bar pressure, is melted and then heated to 298.2 K. The enthalpy of fusion of ice is 6.009 kJ/mole and the heat molar heat capacity of water at 1.000 atm is 75.43 J/ K mole Please help! Can't figure this homework problem out.
Calculate the change in entropy that occurs when 18.02 g of ice at –17.5°C is placed in 90.08 g of water at 100.0°C in a perfectly insulated vessel. Assume that the molar heat capacities for H2O(s) and H2O(l) are 37.5 J K^-1 mol^-1 and 75.3 J K^-1 mol^-1, respectively, and the molar enthalpy of fusion for ice is 6.01 kJ/mol. Change in entropy = ______J/K