How many calories would it take to transform 2 grams of ice at -12C to liquid water at 85C?
How do u find the answere whats the math...and how do u use the chart?
To calculate the number of calories required to transform 2 grams of ice at -12°C to liquid water at 85°C, you need to consider two different processes: raising the temperature of the ice to its melting point, and then melting it into liquid water.
First, you need to find the amount of energy required to raise the temperature of the ice from -12°C to 0°C. The specific heat capacity of ice is 0.5 calories/gram°C. So, you can use the formula:
Energy (calories) = mass (grams) × specific heat capacity (calories/gram°C) × change in temperature (°C)
Plugging in the values:
Energy = 2 grams × 0.5 calories/gram°C × (0°C - (-12°C)) = 2 grams × 0.5 calories/gram°C × 12°C
Next, you need to find the amount of energy required to melt the ice completely, which is called the heat of fusion. The heat of fusion for water is 79.7 calories/gram. So, you can use the formula:
Energy (calories) = mass (grams) × heat of fusion (calories/gram)
Plugging in the values:
Energy = 2 grams × 79.7 calories/gram
Finally, to find the total energy required, you add up the energy for raising the temperature and the energy for melting:
Total Energy (calories) = Energy for temperature change + Energy for melting
Total Energy = (2 grams × 0.5 calories/gram°C × 12°C) + (2 grams × 79.7 calories/gram)
By substituting the values and calculating the equation, you will find the total number of calories required.
As for using a chart, it depends on what kind of chart you are referring to. If you have a chart that provides specific heat capacities or heat of fusion values for different substances, you can simply look up the values and use them in the formulas mentioned above. The chart will provide pre-determined values that are commonly known within the field of thermodynamics.