How much alcohol must be burned to heat a 50g aluminum cylinder from 20 degrees to 200 degrees?

Given:

T1 = 20 degrees Celsius – the initial temperature of the aluminum cylinder;

T2 = 200 degrees Celsius – the required temperature of the aluminum cylinder;

m = 50 grams = 0.05 kilograms is the mass of the cylinder;

c = 920 J / (kg * C) – specific heat capacity of aluminum;

q = 30 MJ / kg = 30 * 10 ^ 6 J / kg is the specific heat of combustion of alcohol.

It is required to determine m1 (kilogram) – how much alcohol needs to be burned in order to heat the cylinder to the temperature T2.

Let’s find the amount of heat required to heat the cylinder:

Q = c * m * (T2 – T1) = 920 * 0.05 * (200 – 20) = 46 * 180 = 8280 Joules.

Since the condition of the problem is not specified, we believe that there is no energy loss during the combustion of alcohol. Then:

m1 = Q / q = 8280 / (30 * 10 ^ 6) = 276 * 10 ^ -6 kilograms = 0.276 grams.

Answer: the required amount of alcohol will be equal to 0.276 grams.