How many calories are required to convert 17g of ice at 0 degrees Celsius to liquid water at 32 degrees Celsius?

Answer 1

The simple beginning is that the definition of a calorie is "the energy required to raise the temperature of 1g of water 1°C." Therefore, the energy required to raise 17g of water 32°C: 17*32=544 cal. However, the question asked about ice. There is an extra bit of energy required for the change of physical state. The energy required to convert 1 gram of ice at 0°C to liquid water at 0°C is called the "latent heat" and is equal to about 80 cal. To convert 17g of ice, we multiply this together: 17g * 80cal/g = 1360 cal. So, we add this energy required for the change of state to the energy required to raise the listed quantity to the required temperature and we get 544 cal + 1360 cal = 1904 cal, assuming no heat is lost to the environment.

I hope this clarifies some things.

Answer 2

To convert 17g of ice at 0 degrees Celsius to liquid water at 32 degrees Celsius, you need to consider two steps: raising the ice temperature from 0°C to 0°C as water (specific heat capacity of water), and then melting the ice into liquid water (latent heat of fusion for water). The total energy required is approximately 420 calories.