Monday, January 29, 2018

HP Prime and TI-84 Plus CE: Heat Properties of Water

HP Prime and TI-84 Plus CE:  Heat Properties of Water


Introduction

The program HEATH20 calculates three heat properties of liquid water based on its temperature.  The program accepts three measurements for liquid water, Fahrenheit (°F), Celsius (°C), or Kelvin (K).

Formulas Used:

Specific heat capacity of water (amount of heat added or removed from an object to cause a temperature change), measured in J/(g*°C):

C =

4.218 + 3.47*10^-4 *(T – 273)^2, for 233 K ≤ T < 274 K 

4.175 + 1.3*10^-5 * (T – 308)^2 + 1.6 * 10^-8 *(T-308)^4, for 274 K ≤ T ≤ 308 K

(slight adjustment on the limits to allow for 273 K to be calculated using the former formula)

Latent heat of evaporation, measured in kJ/kg: 

V = 1000 * (2.5 * (273.15/T)^(0.167 + 3.67 * 10^-4 *T))

Latent heat of fusion, measured in kJ/kg:

M = 333.5 + 2.05 * T – 0.0105 * T^2


HP Prime Program HEATH20

EXPORT HEATH2O()
BEGIN
// Water Properties
// EWS 2018-01-28
// Seinfeld, Pandis
// Atmospheric Chemistry - 2006

// input of temperature
LOCAL ch,T,C,V,H;
INPUT({T,{ch,{"°F","°C","K"}}},
"Temperature",
{"Temp: ","Unit: "});
IF ch==1 THEN
T:=5/9*(T-32)+273.15; END;
IF ch==2 THEN
T:=T+273.15; END;

// Specific heat (liquid)
CASE
IF T≥233 AND T<274 THEN
C:=4.218+3.47ᴇ−4*(T-273)^2; END;
IF T≥274 AND T≤308 THEN
C:=4.175+1.3ᴇ−5*(T-308)+
1.6ᴇ−8*(T-308)^4; END;
DEFAULT
C:=0; END;

// Latent Heat: Evaporation
V:=(2.5*(273.15/T)^
(0.167+3.67ᴇ−4*T))*1000;

// Latent Heat: Fusion
M:=333.5+2.05*(T-273.15)
-0.0105*(T-273.15)^2;

// Results
PRINT();

PRINT("Temp: "+STRING(T)+" K");

IF C≠0 THEN
PRINT("Spec. Heat-Liquid (kJ/g*K)");
PRINT(STRING(C)+" kJ/(g*K)");
END;

PRINT("Latent Heat - Evaporation:");
PRINT(STRING(V)+" kJ/kg");

PRINT("Latent Heat - Fusion");
PRINT(STRING(M)+" kJ/kg");

END;

TI-84 Plus CE Program HEATH20

"EWS 2018-01-28"
"SEINFELD, PANDIS"
"ATMOSPHERIC CHEMISTRY,2006"

Input "TEMPERATURE: ",T
Menu("TEMP. UNIT","°F",1,"°C",2,"K",3)
Lbl 1
5/9*(T-32)+273.15→T
Goto 3
Lbl 2
T+273.15→T
Goto 3
Lbl 3
0→C
If T≥233 and T<274
Then
4.218+3.74E­4*(T-273)^2→C
End
If T≥274 and T≤308
Then
4.175+1.3E­5*(T-308)^2+1.6E­8*(T-308)^4→C
End
(2.5*(273.15/T)^(.167+3.67E­4*T))*1000→V
333.5+2.05*(T-273.15)-.0105*(T-273.15)^2→M
ClrHome
Disp "TEMP (K):",T
If C≠0
Then
Disp "SPEC. HEAT (KJ/(G*K)",C
End
Disp "LATENT-EVAP. (KJ/KG)",V
Disp "LATENT-FUSION (KJ/KG)",M

Example

0°C (273.15 K):

Results:

Temp:  273.15  K
Specific Heat:  4.218008415 kJ/(g*K)
Latent Heat – Evaporation:  2500 kJ/kg
Latent Heat – Fusion:  333.5 kJ/kg

Source:

John H. Seinfeld, Spyros N. Pandis.  Atmospheric Chemistry and Physics: From Air Pollution to Climate Change, 2nd Edition.  John Wiley & Sons, Inc:  Hoboken, New Jersey. 2006  ISBN-13: 978-0-471-72018-8

Eddie


This blog is property of Edward Shore, 2018.

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