how much energy is required to melt a 20.0lb bag of ice at 0c?
heat of fusion is ... 144 btu/lb
To calculate the energy required to melt a substance, you need to use the equation:
Q = m * ΔHf
where Q is the amount of heat energy, m is the mass of the substance, and ΔHf is the heat of fusion (or melting).
First, convert 20.0 pounds to kilograms:
20.0 lbs * (1 kg / 2.20462 lbs) = 9.07185 kg
Next, use the heat of fusion for ice, which is 334 joules/gram:
ΔHf = 334 J/g
Convert the mass from kilograms to grams:
9.07185 kg * (1000 g / 1 kg) = 9071.85 g
Now, calculate the energy required:
Q = m * ΔHf
Q = 9071.85 g * 334 J/g
Q = 3030932.9 J
Therefore, it would take approximately 3,030,932.9 Joules of energy to melt a 20.0 lb bag of ice at 0°C.
To determine the amount of energy required to melt a 20.0 lb bag of ice at 0°C, we need to use the concept of specific heat and latent heat of fusion.
First, we need to calculate the heat required to raise the temperature of the ice from -17.8°C (the average freezing point of freshwater ice) to 0°C.
The specific heat capacity of ice is 2.09 J/g°C. Since there are 453.6 grams in a pound, we can convert 20.0 lb to grams:
20.0 lb = 20.0 lb * 453.6 g/lb ≈ 9072 g.
Next, we calculate the heat required to raise the temperature of the ice using the formula:
Heat = mass * specific heat * change in temperature
Heat = 9072 g * 2.09 J/g°C * (0°C - (-17.8°C)) = 9072 g * 2.09 J/g°C * 17.8°C.
Now, we need to calculate the heat required to melt the ice at 0°C. The latent heat of fusion for ice is 334 J/g.
Heat = mass * latent heat of fusion
Heat = 9072 g * 334 J/g.
Finally, we add the calculated heat values:
Total Heat = Heat to raise the temperature + Heat to melt the ice
Total Heat = 9072 g * 2.09 J/g°C * 17.8°C + 9072 g * 334 J/g.
By simplifying this equation, we can calculate the total energy required to melt the 20.0 lb bag of ice at 0°C.