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.
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