Calculate Grams of Propane Needed for Hot Water Tank | Thermochemistry Help

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SUMMARY

The discussion focuses on calculating the grams of propane (C3H8) needed to heat water in a 50-gallon hot-water tank from 23.4°C to 65.0°C. The heat capacity of the tank is 23.4 J/°C, and it includes 2.3 kg of copper pipe. The specific heat of water is 4.184 kJ/g°C, and the change in temperature is 41.6°C. The heat required for the system is derived from the equation q = specific heat * mass * change in temperature, leading to the conclusion that careful unit conversion and application of thermochemical principles are essential for accurate calculations.

PREREQUISITES
  • Understanding of specific heat capacity and its application in thermochemistry
  • Familiarity with the combustion reaction of propane and its enthalpy change
  • Ability to convert between units, specifically gallons to grams and kilograms to grams
  • Knowledge of heat transfer principles in closed systems
NEXT STEPS
  • Calculate the total heat required using the formula q = specific heat * mass * change in temperature for water, the tank, and copper pipes
  • Learn about the enthalpy of combustion for propane and how it relates to heating calculations
  • Explore unit conversion techniques between gallons, kilograms, and grams for accurate thermochemical calculations
  • Investigate the impact of heat loss in real-world applications and how to account for it in calculations
USEFUL FOR

Students studying thermochemistry, engineers involved in heating system design, and anyone interested in practical applications of heat transfer and combustion reactions.

Charolastra633
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Thermochemistry please help!

Homework Statement



Not sure how to go about this problem.

Calculate the number of grams of propane (C3H8) required to heat all of the water in a 50-gallon hot-water tank from 23.4 degrees C to 65.0 degrees C if the water tank itself has a heat capacity of 23.4J/ degrees C, and it also uses 2.3 kg of copper pipe.

HELP PLEASE!


Homework Equations



SH of water 4.184kj/g oC

q= specific heat* mass * change in temp

The Attempt at a Solution



Change in temp is 41.6 oC change in H rxn for propane combustion = -2042.804 kj

tried coverting the kg into grams and the 50 gal into grams and plugged in the info for heat lost by tank = heat gained by water (sh*mass*change in t) =-(sh*mass*change in t). I don't know what to use anymore and how to put the problem together because I keep getting ridiculous answers.
 
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Intially, the water, tank and copper pipes are all at the same temperature : 23.4*C.
Finally, they all reach the same temperature: 65.0*C

So first, can you calculate how much heat is required to raise the (water) + (tank) + (pipes) through a temperature difference of 41.6*C?

Remember to be careful with units; Joule is an SI (or MKS) unit, as is kilogram.
 

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