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

[Charolastra633]

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.

 

[Gokul43201]

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.

 

[Blueshift5]

I can provide some guidance on how to approach this problem. First, we need to understand the concept of specific heat capacity, which is the amount of heat required to raise the temperature of a substance by 1 degree Celsius. In this case, we are dealing with water, which has a specific heat capacity of 4.184 J/g°C. This means that for every gram of water, we need 4.184 joules of heat to increase its temperature by 1 degree Celsius.

Next, we need to determine the total mass of water in the 50-gallon hot-water tank. This can be done by converting the volume of water (50 gallons) into grams. One gallon of water is equal to approximately 3.785 liters, and one liter of water has a mass of 1000 grams. Therefore, the total mass of water in the tank is equal to 50 gallons * 3.785 liters/gallon * 1000 grams/liter = 189,250 grams.

Now, using the equation q = m * c * ΔT, we can calculate the amount of heat required to raise the temperature of the water from 23.4°C to 65.0°C. We know the specific heat capacity (c) of water and the change in temperature (ΔT) is 41.6°C, so we can plug those values into the equation along with the mass of water (m) we calculated earlier. This will give us the amount of heat (q) required to heat the water in the tank.

Next, we need to account for the heat lost by the hot-water tank and the copper pipes. The heat lost by the tank and pipes will be equal to the heat gained by the water. This is because of the law of conservation of energy, which states that energy cannot be created or destroyed, only transferred from one form to another. We can calculate the heat lost by the tank and pipes by multiplying their respective heat capacities (23.4 J/°C for the tank and 2.3 kg * 0.385 J/g°C for the copper pipes) by the change in temperature (41.6°C).

Finally, we can set up an equation to solve for the mass of propane needed to provide the required amount of heat. This equation will be:

q = m * Hrxn

Where q is the amount of heat required (calculated earlier), m is the mass