1. How much heat energy is needed to transform 125g of ice at 0 degrees Celsius to warm water at 80 degrees Celsius?
q1 = heat to melt ice at zero C to liquid water at zero C.
q1 = mass ice x heat fusion ice.
q2 = heat to move liquid water from zero C to 80 C.
q2 = mass water x specific heat water x (Tfinal-Tinitial)
Total heat = Q = q1 + q2
To determine the amount of heat energy needed for this transformation, we first need to calculate the energy required to raise the temperature of the ice from 0 degrees Celsius to its melting point, then the energy needed to melt the ice, and finally the energy needed to raise the temperature of the resulting water from its melting point to 80 degrees Celsius.
Let's break it down step by step:
Step 1: Calculating the energy to raise the temperature of the ice to its melting point.
The specific heat capacity of ice is 2.09 Joules per gram per degree Celsius (J/g°C). The equation we can use is Q = m × c × ΔT, where Q is the heat energy, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
Q1 = m × c × ΔT1
Q1 = 125g × 2.09 J/g°C × (0°C - (-10°C))
Q1 = 125g × 2.09 J/g°C × 10°C
Q1 = 2612.5 Joules
Step 2: Calculating the energy to melt the ice.
The heat of fusion for ice is 334 Joules per gram (J/g). The equation we can use is Q = m × Hf, where Q is the heat energy, m is the mass, and Hf is the heat of fusion.
Q2 = m × Hf
Q2 = 125g × 334 J/g
Q2 = 41750 Joules
Step 3: Calculating the energy to raise the temperature of the water from its melting point to 80°C.
The specific heat capacity of water is 4.18 J/g°C. The equation we can use is Q = m × c × ΔT, where Q is the heat energy, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
Q3 = m × c × ΔT3
Q3 = 125g × 4.18 J/g°C × (80°C - 0°C)
Q3 = 125g × 4.18 J/g°C × 80°C
Q3 = 5230 Joules
Step 4: Calculating the total heat energy required.
The total heat energy required is the sum of the energy needed for each step.
Total heat energy = Q1 + Q2 + Q3
Total heat energy = 2612.5 Joules + 41750 Joules + 5230 Joules
Total heat energy = 49692.5 Joules
Therefore, the amount of heat energy needed to transform 125g of ice at 0 degrees Celsius to warm water at 80 degrees Celsius is approximately 49692.5 Joules.
To calculate the amount of heat energy required to transform ice into warm water, we need to consider two processes: the energy required to heat the ice from 0 degrees Celsius to its melting point, and the energy required to further heat the water from its melting point to 80 degrees Celsius.
1. Energy required to heat the ice to its melting point:
The specific heat capacity of ice is 2.09 J/g°C. Therefore, to heat 125g of ice from 0 degrees Celsius to its melting point, use the formula:
Q1 = mass * specific heat capacity * ΔT
Q1 = 125g * 2.09 J/g°C * (0°C - 0°C)
Q1 = 0 J
Since there is no temperature change, no energy is required to heat the ice to its melting point.
2. Energy required to melt the ice:
The enthalpy of fusion for ice is 334 J/g. Therefore, to melt 125g of ice, use the formula:
Q2 = mass * enthalpy of fusion
Q2 = 125g * 334 J/g
Q2 = 41800 J
3. Energy required to heat the water from its melting point to 80 degrees Celsius:
The specific heat capacity of water is 4.18 J/g°C. Therefore, to heat 125g of water from its melting point to 80 degrees Celsius, use the formula:
Q3 = mass * specific heat capacity * ΔT
Q3 = 125g * 4.18 J/g°C * (80°C - 0°C)
Q3 = 523750 J
To find the total energy required, sum up the energy from each step:
Total energy = Q1 + Q2 + Q3
Total energy = 0 J + 41800 J + 523750 J
Total energy = 565550 J
Therefore, the amount of heat energy needed is 565550 Joules.