How many grams of methane must be combusted to heat 1.05 of water from 24.0 to 89.0, assuming as a product and 100% efficiency in heat transfer?

To find out how many grams of methane must be combusted, we need to calculate the amount of heat transferred to the water. The equation we will use is:

q = mcΔT

Where:
q = heat transferred (in joules)
m = mass of water (in grams)
c = specific heat capacity of water (4.18 J/g°C)
ΔT = change in temperature (in °C)

First, let's calculate the mass of water:
mass of water (m) = 1.05 kg * 1000 g/kg = 1050 g

Next, let's calculate the change in temperature:
ΔT = final temperature - initial temperature = 89.0°C - 24.0°C = 65.0°C

Now, we can use the equation to find the heat transferred:
q = mcΔT
q = 1050 g * 4.18 J/g°C * 65.0°C

Calculating the heat transferred gives us the total energy needed to heat the water. Assuming 100% efficiency in heat transfer, this will be the same amount of energy produced by combusting methane.

Now, we need to convert the energy to joules to calculate the amount of methane required. Assume that the combustion of methane releases 802 kJ per mole of methane.

1 J = 1 kJ/1000

q = 1050 g * 4.18 J/g°C * 65.0°C * (1 kJ/1000 J)

Next, we need to find the number of moles of methane required using the energy conversion:

1 mole of methane = 802 kJ

moles of methane = q / 802 kJ/mole

Finally, we can calculate the mass of methane using its molar mass, assuming that methane has a molar mass of 16.04 g/mol.

grams of methane = moles of methane * 16.04 g/mol

Calculating these values will give you the answer to your question.