How many cans ( each containing 14.0 oz of propane) would be required to convert 41.7 lbs of ice at -15.0 C to water at 100 C ( the boiling point at standard pressure)
To determine the number of cans required to convert 41.7 lbs of ice at -15.0 °C to water at 100 °C, we need to calculate the amount of heat energy required to melt the ice and then heat the resulting water to its boiling point.
Step 1: Calculate the heat energy required to melt the ice:
The heat energy required to melt ice can be calculated using the formula:
Q = m * heat fusion, where
Q is the heat energy required,
m is the mass of ice, and
heat fusion is the heat of fusion.
Given:
Mass of ice, m = 41.7 lbs = 18.9 kg (since 1 lb = 0.4536 kg)
Heat of fusion for ice = 334 J/g (joules per gram)
First, convert the mass from kg to grams:
m = 18.9 kg * 1000 = 18,900 g
Now, calculate the heat energy required to melt the ice:
Q1 = m * heat fusion
Q1 = 18,900 g * 334 J/g = 6,313,400 J
Step 2: Calculate the heat energy required to heat the water from -15.0 °C to 100 °C:
The heat energy required to heat a substance can be calculated using the formula:
Q = m * specific heat capacity * ΔT, where
Q is the heat energy required,
m is the mass of the substance,
specific heat capacity is the amount of heat energy required to raise the temperature of the substance by 1 degree, and
ΔT is the change in temperature.
Given:
Mass of ice (now melted), m = 18.9 kg
Specific heat capacity of water = 4.18 J/g°C
Change in temperature, ΔT = (100 °C) - (-15.0 °C) = 115.0 °C
First, convert the mass from kg to grams:
m = 18.9 kg * 1000 = 18,900 g
Now, calculate the heat energy required to heat the water:
Q2 = m * specific heat capacity * ΔT
Q2 = 18,900 g * 4.18 J/g°C * 115.0 °C = 9,386,610 J
Step 3: Calculate the total heat energy required:
The total heat energy required is the sum of the heat energy required to melt the ice (Q1) and the heat energy required to heat the water (Q2):
Total heat energy required = Q1 + Q2
Total heat energy required = 6,313,400 J + 9,386,610 J = 15,700,010 J
Step 4: Calculate the number of propane cans required:
Each can contains 14.0 oz (ounces) of propane. To find the number of cans, we need to convert the total heat energy required (in joules) to the heat energy produced by one can of propane.
The heat energy produced by one can of propane = 14.0 oz * (28.35 g/oz) * heat of combustion of propane
The heat of combustion of propane is approximately 50 kJ/g (kilojoules per gram).
Convert the heat energy produced by one can of propane to joules:
Heat energy produced by one can of propane = (14.0 oz * 28.35 g/oz) * 50,000 J/g = 19,845,000 J
Finally, calculate the number of cans required:
Number of cans required = Total heat energy required / heat energy produced by one can of propane
Number of cans required = 15,700,010 J / 19,845,000 J ≈ 0.79 cans
Therefore, approximately 0.79 cans (or 1 can) of propane would be required to convert 41.7 lbs of ice at -15.0 °C to water at 100 °C.