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

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To calculate the grams of propane needed to heat a 50-gallon hot water tank from 23.4°C to 65.0°C, the total heat required must be determined, considering the specific heat of water, the heat capacity of the tank, and the mass of the copper pipes. The specific heat of water is 4.184 kJ/g°C, and the heat gained by the water must equal the heat lost by the propane combustion, which has a change in enthalpy of -2042.804 kJ. The temperature change is 41.6°C, and the calculations should include converting the water volume and mass of the pipes into appropriate units. Careful attention to units is crucial for accurate results. The problem requires integrating these elements to find the total grams of propane needed.
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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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