How much butane would you need to make a cup (250mL) of instant hot chocolate from a very cold stream (1.8 degree Celsius) if you wanted to heat the water to 90 degree Celsius.

Answer 1

To calculate the amount of butane required to heat the water from 1.8 degree Celsius to 90 degree Celsius, we need to consider the heat capacity of water.

The specific heat capacity of water is approximately 4.18 J/g°C, which means it takes 4.18 Joules of energy to raise the temperature of 1 gram of water by 1 degree Celsius.

First, we need to determine the mass of the water:

250 mL of water is equal to 250 grams (since the density of water is roughly 1 g/mL).

Next, we calculate the temperature difference:

Temperature difference = final temperature - initial temperature
Temperature difference = 90°C - 1.8°C
Temperature difference = 88.2°C

Now, we can calculate the energy required to heat the water from 1.8°C to 90°C:

Energy = mass of water × specific heat capacity × temperature difference
Energy = 250 g × 4.18 J/g°C × 88.2°C
Energy ≈ 92,355 Joules

Butane has a lower heating value of approximately 45,000 kJ/kg, which is equal to 45,000,000 J/kg. Since the amount of butane is not given, we need to assume a certain efficiency of heat transfer to determine the minimum amount of butane required.

For this calculation, let's assume an efficiency of 50% for the heat transfer:

Minimum amount of butane required = Energy required / efficiency
Minimum amount of butane required = 92,355 J / (0.50)
Minimum amount of butane required = 184,710 Joules

Converting the energy value to kilograms:

Number of kilograms of butane required = Minimum amount of butane required / lower heating value
Number of kilograms of butane required = 184,710 J / 45,000,000 J/kg
Number of kilograms of butane required ≈ 0.0041 kg

Therefore, you would need approximately 0.0041 kilograms (or 4.1 grams) of butane to heat 250 mL of water from 1.8°C to 90°C, assuming an efficiency of 50% for the heat transfer.