Calculate the amount of heat in kJ that is required to heat 25.0 g of ice from -25 °C to 105 °C in a closed vessel and sketch a
heating curve for the process. The specific heat of ice is 2.11 J/(g. "C); 4.18 J/g. "C) for water, 2.00 J/g. "C. AHus for water is 6,01 kJ/mol; AHp for water = 40.67 kJ/mol.
1 answer:
Answer:
The total amount of heat required for the process is 76.86 KJ
Explanation:
We can divide the process in 5 parts, in which we can calcule each amount of heat required (see attached Heating curve):
(1) Ice is heated from -25ºC to 0ºC. We can calculate the heat of this part of the process as follows. Note that we must convert J in KJ (1 KJ= 1000 J).
Heat (1) = mass ice x Specific heat ice x (Final temperature - Initial Temperature)
Heat (1) =
Heat (1) = 1.32 KJ
(2) Ice melts at ºC (it becomes liquid water). This is heating at constant temperature (ºC), so we use the melting enthalphy (ΔHmelt) and we must use the molecular weight of water (1 mol H₂O = 18 g):
Heat (2) = mass ice x ΔHmelt
Heat (2)=
Heat (2)= 8.35 KJ
(3) Liquid water is heated from 0ºC to 100 ºC:
Heat (3)= mass liquid water x Specific heat water x (Final T - Initial T)
Heat (3)= 25 g x 4.18 J/gºC x 1 KJ/1000 J x (100ºC - 0ºC)
Heat (3)= 10.45 KJ
(4) Liquid water evaporates at 100ºC (it becomes water vapor). This is a process at constant temperature (100ºC), and we use boiling enthalpy:
Heat (4)= mass water x ΔH boiling
Heat (4)= 25 g x x
Heat (4)= 56.49 KJ
(5) Water vapor is heated from 100ºC to 105ºC. We use the specific capacity of water vapor:
Heat (5)= mass water vapor x Specific capacity vapor x (Final T - Initial T)
Heat (5)= 25 g x 2.00 J/g ºC x 1 KJ/1000 J x (105ºC - 100ºC)
Heat (5)= 0.25 KJ
Finally, we calculate the total heat involved in the overall process:
Total heat= Heat(1) + (Heat(2) + Heat(3) + Heat(4) + Heat(5)
Total heat= 1.32 KJ + 8.35 KJ + 10.45 KJ + 56.49 KJ + 0.25 KJ
Total heat= 76.86 KJ
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