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.
To calculate the amount of butane needed to heat the water for making a cup of instant hot chocolate, we can use the specific heat capacity formula:
Q = mcΔT
Where:
Q = heat energy (in joules)
m = mass of the water (in grams)
c = specific heat capacity of water (4.18 J/g°C)
ΔT = change in temperature (final temperature - initial temperature)
First, convert the volume of water from milliliters to grams. Assuming the density of water is 1 g/mL, the mass of 250 mL of water is 250 grams.
Next, calculate the change in temperature (ΔT):
ΔT = 90°C - 1.8°C = 88.2°C
Now we can calculate the heat energy (Q) required to heat the water:
Q = (250 g) * (4.18 J/g°C) * (88.2°C)
Q = 92,214 J
Assuming that the combustion of butane is 100% efficient and that the heat energy released by butane is transferred to the water, the amount of butane needed can be calculated using the heat of combustion. The heat of combustion for butane is approximately 2878 kJ/mol, which is equivalent to 2878 x 10^3 J/mol.
The molar mass of butane is approximately 58.12 g/mol.
To find the moles of butane needed, we can divide the heat energy required by the heat of combustion of butane:
Moles of butane = (92,214 J) / (2878 x 10^3 J/mol)
Moles of butane = 0.032 mol
Since the molar mass of butane is 58.12 g/mol, we can calculate the mass of butane needed:
Mass of butane = (0.032 mol) * (58.12 g/mol)
Mass of butane = 1.86 g
Therefore, you would need approximately 1.86 grams of butane to heat 250 mL of water from 1.8°C to 90°C.
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.
To calculate the amount of butane needed to heat the water, we need to consider the specific heat capacity of water and the heat of combustion of butane.
1. Determine the temperature change:
The temperature needs to increase from 1.8 °C to 90 °C.
ΔT = final temperature - initial temperature
ΔT = 90 °C - 1.8 °C
ΔT = 88.2 °C
2. Calculate the heat energy required:
The heat energy required can be calculated using the formula:
Q = m * c * ΔT
where Q is the heat energy (in Joules), m is the mass of water (in grams), c is the specific heat capacity of water (4.18 J/g°C), and ΔT is the temperature change (in °C).
Convert the volume of water into mass by multiplying it by the density of water (1 g/mL):
Mass of water = volume * density
Mass of water = 250 mL * 1 g/mL
Mass of water = 250 grams
Now substitute the values into the formula:
Q = m * c * ΔT
Q = 250 g * 4.18 J/g°C * 88.2 °C
Q ≈ 92,435 J
3. Convert the heat energy to amount of butane:
The heat of combustion of butane is approximately 50.3 kJ/g or 50,300 J/g.
To find the mass of butane needed, divide the heat energy required by the heat of combustion of butane:
Mass of butane = Q / heat of combustion of butane
Mass of butane = 92,435 J / 50,300 J/g
Mass of butane ≈ 1.84 grams
Therefore, you would need approximately 1.84 grams of butane to heat 250 mL of water from 1.8 °C to 90 °C.