Question

how much methanol is required to heat a 48 g. aluminum pot holding 400 ml. of water

Answer 1

The mass of methanol required to heat a 48 g aluminum pot holding 400 ml of water is 28.5 g.

What is heat energy change?

The heat energy change q is given below as follows;

• q = mcΔT

where;

• m is mass
• c is the specific heat capacity

• ΔT is the temperature difference

Data given:

mass of Al = 48 g

ΔT= 40 -25

ΔT = 15° C

specific heat of Al = 0.900 J/g° C

q = 48 g × 0.900 J/g° C × 15° C

q =  648 J

the amount of energy transferred is 648 J

b. The amount of methanol required is calculated as follows:

The molar enthalpy of methanol is -727 KJ/mol.

The moles of ethanol required = 648 J / 727 KJ/mol

The moles of ethanol required = 0.891 moles

the molar mass of ethanol = 32 g/mol

mass of methanol = 0.891 * 32 g

mass of methanol required = 28.5 g

Learn more about heat energy change at: https://brainly.com/question/24298104

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Complete question:

1. A 48 g. aluminum pot holding 400 ml. of water is heated from 25 to 40 degrees Celcius.

a. Calculate the amount of energy transferred. Note Aluminum has a specific heat capacity of 0.900 J/g C.

b) If the heat was transferred by a methanol burner with molar enthalpy of -727 KJ/mol., calculate the amount (mass in grams) of methanol required to heat the water and the aluminum.