How do you find final temperature

Question

The molar enthalpy of combustion for methane (CH4) is –2242.0 J/mol. 0.50 g of methane is used to heat 45.0 mL of water at 24.0°C. What is the final temperature of the water (assuming perfect heat transference)?

Answer 1

The heat released by burning 0.50 g of methane (CH4) will be used to heat the 45.0 ml of water, so the temperature of the water will, of course, rise.

Step 1: Find the heat generated by burning 0.50 g of CH4:

0.50 g CH4 x 1 mole CH4 / 16 g = 0.03125 moles CH4

0.03125 mols CH4 x -2242.0 J/mol = -70.06 J (negative because heat is being released)

Step 2: Determine the final temperature of the water after absorbing 70.06 J of heat energy:

For this calculation, we use q = mC∆T

q = heat = 70.06 J

m = mass of water = 45.0 ml x 1 g / ml = 45.0 g (assuming a density of 1g/ml for water)

C = specific heat for water = 4.184 J/g/º (this is a constant that you can look up)

∆T = change in temperature = ?

Solving for ∆T, we have...

70.06 J = (45.0 g)(4.184 J/gº)(∆T)

∆T = 0.37º

Since the temperature of the water will increase, we simply add this to the original temperature of the water to find the final temperature:

0.37º + 24º = 24.37º = 24.4ºC = FINAL TEMPERATURE

Answer 2

For this kind of question, it is important to remember the heat released by burning of methane is equal to the heat absorbed by the water.

First calculate the moles of methane: 0.5 g/ [(12+4)g/ mol] = 0.03125 mol

Second calculate the heat released by burning of methane: △H = -2242.0 J/mol * 0.03125 mol = -70.0625 J

Third assume the final temperature of water is Tf, thus the heat absorbed by the water is △H = c*m*△T = c*m*(Tf-24°C).

The density of water is 1 g/ mL, thus the mass of 45 mL of water is 45 g

The specific heat capacity of water is 4.184J/(g*°C), thus he heat absorbed by the water is △H = 4.184J/(g*°C) * 45 g *(Tf- 24°C), which is equal to the heat released by burning of 0.5 g of methane (Notice: the numbers of the heat released and absorbed are equal but the value of released heat is negative and the value of absorbed heat is positive).

Therefore, 4.184J/(g*°C) * 45 g *(Tf- 24°C) = 70.0625 J.

Tf = 24.4 °C

The final temperature of the water is 24.4°C.