calculate the amount of heat required to change 5 pounds of ice at 32 degrees to water 32 degrees
5 lbs x 144 Btu / lbs = 720 Btu
Well, well, well, looks like we have some frozen water on our hands. Let's do a little heat calculation dance!
To change ice at 32 degrees Fahrenheit to water at the same temperature, we need to heat it up to its melting point. And to do that, we use the magical number known as the heat of fusion.
The heat of fusion for water is 144 BTU (British Thermal Units) per pound. And since we have 5 pounds of ice, we can put on our math hats and start crunching some numbers.
5 pounds of ice * 144 BTU/pound = 720 BTU
So, there you have it! You'll need a roaring 720 BTUs of heat to turn that ice into water. Just be careful not to slip and land on your funny bone!
To calculate the amount of heat required to change ice at 32 degrees Fahrenheit to water at 32 degrees Fahrenheit, we need to consider the three steps involved in this process:
1. Heating the ice from its initial temperature of 32 degrees Fahrenheit to 32 degrees Fahrenheit.
2. Melting the ice into water at 32 degrees Fahrenheit.
3. Heating the water from 32 degrees Fahrenheit to 32 degrees Fahrenheit.
Let's calculate the amount of heat required for each step:
1. Heating the ice from 32°F to 32°F:
The specific heat capacity of ice is 0.5 calories/gram°C, which is approximately 1 calorie/gram°F. Since there are 453.6 grams in a pound, the heat required to heat the ice can be calculated as follows:
Q1 = (5 pounds) x (453.6 grams/pound) x (1 calorie/gram°F) x (32°F - 32°F) = 0 calories
2. Melting the ice into water at 32°F:
The heat of fusion for ice is 333.5 J/g or 79.7 calories/g. Therefore, the heat required to melt the ice can be calculated as:
Q2 = (5 pounds) x (453.6 grams/pound) x (79.7 calories/gram) = 181,620 calories
3. Heating the water from 32°F to 32°F:
The specific heat capacity of water is 1 calorie/gram°C, which is approximately 1 calorie/gram°F. So, the heat required to heat the water can be calculated as:
Q3 = (5 pounds) x (453.6 grams/pound) x (1 calorie/gram°F) x (32°F - 32°F) = 0 calories
The total amount of heat required is the sum of Q1, Q2, and Q3:
Total heat required = Q1 + Q2 + Q3 = 0 calories + 181,620 calories + 0 calories = 181,620 calories
Therefore, the amount of heat required to change 5 pounds of ice at 32 degrees Fahrenheit to water at 32 degrees Fahrenheit is 181,620 calories.
To calculate the amount of heat required to change 5 pounds of ice at 32 degrees Fahrenheit to water at 32 degrees Fahrenheit, we need to consider the heat involved in each phase change.
1. Heating the ice from -32 degrees Fahrenheit to 32 degrees Fahrenheit:
- The formula to calculate the heat required to change the temperature of a substance is Q = m * c * ΔT, where Q is the heat energy, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.
- The specific heat capacity of ice is approximately 0.5 calories/gram °C.
- First, we need to convert pounds to grams. Since 1 pound is approximately 453.6 grams, we have:
Mass of ice = 5 pounds * 453.6 grams/pound = 2268 grams.
- The change in temperature is 32 degrees Fahrenheit - (-32) degrees Fahrenheit = 64 degrees Fahrenheit = 17.8 degrees Celsius.
- Now we can calculate the heat required to heat the ice from -32 degrees Fahrenheit to 32 degrees Fahrenheit:
Q = 2268 grams * 0.5 calories/gram °C * 17.8 °C.
2. Melting the ice to water at 32 degrees Fahrenheit:
- The heat required to melt ice is given by the formula Q = m * ΔH, where m is the mass of the substance and ΔH is the enthalpy of fusion.
- The enthalpy of fusion of ice is approximately 333.55 Joules/gram or 79.7 calories/gram.
- The mass of the ice remains the same, which is 2268 grams.
- The heat required to melt the ice is therefore:
Q = 2268 grams * 79.7 calories/gram.
3. Heating the water from 32 degrees Fahrenheit to 32 degrees Fahrenheit:
- The specific heat capacity of water is approximately 1 calorie/gram °C.
- The mass of water is also 2268 grams.
- The change in temperature is 32 degrees Fahrenheit - 32 degrees Fahrenheit = 0 degrees Fahrenheit = 0 degrees Celsius.
- The heat required to heat the water is therefore:
Q = 2268 grams * 1 calorie/gram °C * 0 °C.
To find the total heat required to change 5 pounds of ice at 32 degrees Fahrenheit to water at 32 degrees Fahrenheit, we need to add up the heat for each step:
Total Heat = Heat to heat ice + Heat to melt ice + Heat to heat water
Note: You can convert the result from calories to other units like Joules or BTUs if needed.
q to melt ice
mass ice(in grams) x heat fusion.
Is that 32 F or 32 C. If 32 C, that will be
q = mass melted ice x specific heat x (Tfinal-Tinitial)
Then add the two qs together.