How much heat energy is needed to transform 125g of ice at zero degrees celsius to warm water at 80 degrees celsius?
I answered this for you yesterday.
whta's the point of this website of no ones gonna answer the frickin question????????????
To determine the amount of heat energy needed for the transformation described, we need to consider the following steps:
1. Heat energy required to raise the temperature of the ice from -273.15°C to 0°C
2. Heat energy required to melt the ice at 0°C into liquid water at 0°C
3. Heat energy required to raise the temperature of the liquid water from 0°C to 80°C
Let's calculate each step separately:
Step 1: Heating the ice to 0°C.
The specific heat capacity of ice is 2.09 J/g°C.
To calculate the heat energy required, we use the formula:
Q1 = mass x specific heat capacity x change in temperature
Q1 = 125g x 2.09 J/g°C x (0°C - (-273.15°C))
Q1 = 125g x 2.09 J/g°C x 273.15°C
Q1 = 71,025.94 J
Step 2: Melting the ice at 0°C into liquid water at 0°C.
The heat of fusion (enthalpy of fusion or latent heat) of ice is 334.6 J/g.
To calculate the heat energy required, we use the formula:
Q2 = mass x heat of fusion
Q2 = 125g x 334.6 J/g
Q2 = 41,825 J
Step 3: Heating the water from 0°C to 80°C.
The specific heat capacity of water is 4.18 J/g°C.
To calculate the heat energy required, we use the formula:
Q3 = mass x specific heat capacity x change in temperature
Q3 = 125g x 4.18 J/g°C x (80°C - 0°C)
Q3 = 52,425 J
Finally, to get the total heat energy required, we sum up the heat energy calculated in each step:
Total heat energy = Q1 + Q2 + Q3
Total heat energy = 71,025.94 J + 41,825 J + 52,425 J
Total heat energy ≈ 165,276 J
Therefore, approximately 165,276 Joules of heat energy are needed to transform 125g of ice at zero degrees Celsius to warm water at 80 degrees Celsius.