1. How much heat energy is needed to transform 125g of ice at 0 degrees Celsius to warm water at 80 degrees Celsius?
See your post above.
To find the amount of heat energy required to transform ice into warm water, we need to consider two steps:
1. Heat energy needed to raise the temperature of the ice from 0 degrees Celsius to its melting point (0 degrees Celsius).
2. Heat energy needed to melt the ice and raise the temperature of the resulting water from its melting point to 80 degrees Celsius.
Let's calculate each step using specific heat capacities and heat of fusion:
Step 1: Heating the ice from 0 degrees Celsius to its melting point
The specific heat capacity of ice is 2.09 J/g°C, which means it takes 2.09 Joules of heat energy to raise the temperature of 1 gram of ice by 1 degree Celsius.
To find the heat energy needed to raise the temperature of 125 grams of ice from 0 degrees Celsius to 0 degrees Celsius (its melting point), we can use the formula:
Q = m * c * ΔT
where:
Q = heat energy (in joules)
m = mass of the substance (in grams)
c = specific heat capacity (in J/g°C)
ΔT = change in temperature (in °C)
In this case:
m = 125 grams
c = 2.09 J/g°C
ΔT = 0°C - 0°C = 0°C
Q1 = 125g * 2.09 J/g°C * 0°C = 0 J
So, no heat energy is required for this step since there is no temperature change.
Step 2: Melting the ice and heating the resulting water from its melting point to 80 degrees Celsius
The heat of fusion for ice is 334 J/g, which means it takes 334 Joules of heat energy to melt 1 gram of ice.
To find the heat energy needed to melt 125 grams of ice, we can use the formula:
Q = m * Hf
where:
Q = heat energy (in joules)
m = mass of the substance (in grams)
Hf = heat of fusion (in J/g)
In this case:
m = 125 grams
Hf = 334 J/g
Q2 = 125g * 334 J/g = 41750 J
So, 41750 Joules of heat energy are required to melt the ice.
Next, we need to find the heat energy needed to raise the temperature of the resulting water from its melting point (0 degrees Celsius) to 80 degrees Celsius. The specific heat capacity of water is 4.18 J/g°C.
Again, using the formula Q = m * c * ΔT:
Q3 = 125g * 4.18 J/g°C * (80°C - 0°C) = 523750 J
Therefore, the total heat energy required to transform 125g of ice at 0 degrees Celsius to warm water at 80 degrees Celsius is:
Total Q = Q1 + Q2 + Q3 = 0 J + 41750 J + 523750 J = 565500 J
So, 565500 Joules of heat energy are needed.