How many joules are needed to heat 8.50 grams of ice from -10.0 degrees C to 25.0 degrees C?
q1 = heat necessary to change ice from -10 to zero = mass x specific heat ice x (Tfinal - Tinitial) where Tfinal is zero and Tinitial is -10.
q2 = heat necessary to melt ice but stay at zero = mass x heat fusion of ice.
q3= heat necessary to heat water at zero to 25 C. = mass x specific heat water x (25-0).
total = q1 + q2 + q3.
Post your work if you get stuck.
To determine the amount of energy required to heat the ice, we need to consider the heat energy required at different stages:
1. Heating the ice from -10.0°C to 0°C: Since ice is at a temperature below its melting point, we need to calculate the energy required to heat it to 0°C. For this step, we'll use the formula:
Q = mcΔT
Where:
Q is the heat energy
m is the mass of the ice
c is the specific heat capacity of ice (2.09 J/g°C)
ΔT is the change in temperature (0°C - (-10.0°C) = 10.0°C)
Plugging in the values, we have:
Q = (8.50 g) * (2.09 J/g°C) * (10.0°C)
Q = 178.85 J
2. Melting the ice at 0°C: The heat energy required to convert the ice to water at its melting point is given by:
Q = mL
Where:
Q is the heat energy
m is the mass of the ice
L is the heat of fusion (334 J/g)
Plugging in the values, we have:
Q = (8.50 g) * (334 J/g)
Q = 2839 J
3. Heating the water from 0°C to 25.0°C: Now that the ice has melted into water, we need to calculate the energy required to heat the water. For this step, we'll use the same formula as in step 1:
Q = mcΔT
Where:
Q is the heat energy
m is the mass of the water (which is equivalent to the mass of the ice)
c is the specific heat capacity of water (4.18 J/g°C)
ΔT is the change in temperature (25.0°C - 0°C = 25.0°C)
Plugging in the values, we have:
Q = (8.50 g) * (4.18 J/g°C) * (25.0°C)
Q = 888.25 J
Adding up the energies from each step, we get:
Total energy = 178.85 J + 2839 J + 888.25 J = 3906.1 J
Therefore, it would require approximately 3906.1 joules of energy to heat 8.50 grams of ice from -10.0 degrees Celsius to 25.0 degrees Celsius.