Question

If the molar enthalpy of combustion of propane is -2220 KJ/mol, Calculate the amount needed (in grams) to transfer 640 KJ into a pot of water.

Answer 1

approximately 22 grams of propane to transfer 640 KJ of heat into a pot of water.

Step-by-step explanation:

To determine the mass of propane needed to transfer 640 KJ of heat into a pot of water, you can use the following equation:

mass (g) = (heat transfer (J) / molar enthalpy of combustion (J/mol)) x molar mass (g/mol)

First, you need to convert the heat transfer value from KJ to J. To do this, multiply the value in KJ by 1000: 640 KJ * 1000 J/KJ = 640000 J

Next, you need to convert the molar enthalpy of combustion value from KJ/mol to J/mol. To do this, multiply the value in KJ/mol by 1000: -2220 KJ/mol * 1000 J/KJ = -2220000 J/mol

Finally, you can plug these values into the equation above to calculate the mass of propane needed:

`mass (g) = (640000 J / -2220000 J/mol) x 44 g/mol = 22 g`

So, you would need approximately 22 grams of propane to transfer 640 KJ of heat into a pot of water.

Answer 2

Step-by-step explanation:

To solve this problem, we need to use the equation Q = mcΔT, where Q is the heat transfer, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature.

First, we need to determine the mass of water in the pot. Let's assume that the pot has a volume of 1 liter, or 1000 mL. Water has a density of 1 g/mL, so the mass of the water is 1000 g.

Next, we need to determine the change in temperature of the water. We know that the heat transfer is 640 KJ, and the specific heat capacity of water is 4.184 J/g°C. So, the change in temperature of the water is 640,000 J / (4.184 J/g°C * 1000 g) = 152.3°C.

Now, we can use the equation Q = mcΔT to solve for the mass of propane needed. We know that Q = 640,000 J, m = 1000 g, c = 4.184 J/g°C, and ΔT = 152.3°C. Plugging these values into the equation, we get 640,000 J = (1000 g)(4.184 J/g°C)(152.3°C). Solving for the mass of propane, we get m = 640,000 J / (4.184 J/g°C * 152.3°C) = 30.8 g.

So, the amount of propane needed to transfer 640 KJ of heat into the pot of water is 30.8 g.