How many Joules are needed to melt 1 gram of 0 deg C ice and turn it to water at room temperature of 23 deg C?

429 j

q1 to melt ice = 1g x 334 J/g = 334 J

q2 to heat water to 23°C from 0°C:
- Mass of water = 1g
- Specific heat of water = 4.184 J/(g·°C)
- Final temperature = 23°C
- Initial temperature = 0°C
q2 = 1g x 4.184 J/(g·°C) x (23°C - 0°C) = 97.192 J
Qtotal = q1 + q2 = 334 J + 97.192 J = 431.192 J

Therefore, approximately 431 Joules are needed to melt 1 gram of 0°C ice and turn it into liquid water at 23°C.

Ah, the joys of melting ice! Well, to answer your question, the amount of energy required to melt ice is determined by its specific heat capacity and heat of fusion.

The specific heat capacity of ice is approximately 2.09 J/g°C, and the heat of fusion is about 334 J/g. So, let's do some fun math!

First, we calculate the energy needed to raise the temperature of ice from 0°C to 0°C (yes, I know, it sounds redundant). Since there's no change in temperature, no energy is needed here.

Next, to melt the ice, we multiply the heat of fusion by the mass of ice:

334 J/g * 1 g = 334 J

Now that we have melted the ice, we need to raise the temperature of the resulting water from 0°C to 23°C. Using the specific heat capacity of water (about 4.18 J/g°C), we calculate:

4.18 J/g°C * 1 g * 23°C = 96.14 J

Adding up the energy needed to melt the ice and raise the temperature of the resulting water, we get:

334 J + 96.14 J = 430.14 J

So, approximately 430.14 Joules of energy are needed to melt 1 gram of ice at 0°C and raise it to water at 23°C. Quite chilly, huh?

To calculate the total amount of energy required to melt the ice and heat it to room temperature, we need to consider two separate steps:

Step 1: Heat energy required to raise the temperature of 1 gram of ice from 0°C to 0°C (melting point)
Step 2: Heat energy required to raise the temperature of 1 gram of water from 0°C to 23°C (room temperature)

Step 1: Melting the ice
The heat energy required to melt the ice is known as the latent heat of fusion. For ice, the latent heat of fusion is approximately 334 joules per gram.

So, to calculate the amount of energy required to melt 1 gram of ice, we multiply the mass (1 gram) by the latent heat of fusion (334 J/g):
Energy = Mass x Latent heat of fusion
Energy = 1 gram x 334 J/g
Energy = 334 Joules

Step 2: Heating the water from 0°C to 23°C
The specific heat capacity of water is approximately 4.18 joules per gram per degree Celsius. This means that it takes 4.18 joules to raise the temperature of 1 gram of water by 1 degree Celsius.

To raise the temperature of 1 gram of water from 0°C to 23°C, we need to calculate the energy required for a temperature increase of 23°C:
Energy = Mass x Specific heat capacity x Temperature change
Energy = 1 gram x 4.18 J/g°C x 23°C
Energy = 96.14 Joules

To get the total energy required, we add the energy from Step 1 to the energy from Step 2:
Total Energy = Energy for melting + Energy for heating
Total Energy = 334 Joules + 96.14 Joules
Total Energy ≈ 430.14 Joules

Therefore, approximately 430.14 Joules are needed to melt 1 gram of ice at 0°C and heat it to water at room temperature (23°C).

q1 to melt ice = 1g x heat fusion in J/g = ?

q2 to heat water to 23 C.frin zero C
q2 = mass water x specific heat H2O x (Tfinal-Tinitial)
Qtotal q1 + q2 = ?