A typical bathtub can hold 93 gallons of

Question

water. Calculate the mass of natural gas that would need to be burned to heat the water for a tub of this size from 60◦F to 99◦F. Assume that the natural gas is pure methane(CH4) and that the products of combustion are carbon dioxide and water (liquid).
Answer in units of g

Answer 1

This is a time burner. What do you not understand about this?

Answer 2

i don't know equation to use.

Answer 3

Convert 93 gallons H2O to grams. I would use 1 gallon = 3.785 L, convert to mL then use density of 1.00 g/mL to solve for grams. How much heat is required to heat this much water (in grams) from 60 F to 99 F. Convert 60 F and 99 F to C using (F-32)*5/9 = C.
Then q = heat required = mass H2O x specific heat H2O (that's 4.184 J/g*C) x (Tfinal-Tinitial) with T in C (not in F)

Now that you know how much heat is required, you want to calculate the amount of CH4 needed to produce that heat. If you don't know the heat of combustion you can calculate it this way.
CH4 + 2O2 ==> CO2 + 2H2O
dHrxn = (n*dH products) - (n*dH reactants). That is about 891 kJ/mol CH4 (note H2O is a liquid here).

So you need 16g x (q from above/891)= ?g CH4.

Answer 4

Okay. I did each of these steps and I didn't get the correct answer.

Answer 5

What volume of natural gas does this correspond to at 25◦C and 0.8 atm?
Answer in units of L
I could not find the answer to the question above but this question uses the answer to the other question. So can you please help me.