Final answer:
To calculate the mass of lead required to generate 3000 joules of energy when cooled from 100 °C to 40 °C, we can use the formula E = mcΔT, where E is the energy, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. Knowing the specific heat capacity of aluminum, we can use it as an approximation for lead. Substituting the known values into the formula, we find that it would take approximately 0.0556 kilograms of lead to generate 3000 joules of energy.
Step-by-step explanation:
To calculate the amount of energy needed to change the temperature of a substance, we use the formula:
E = mcΔT
Where:
- E is the energy in joules
- m is the mass of the substance in kilograms
- c is the specific heat capacity of the substance in J/kg-°C
- ΔT is the change in temperature in °C
In this case, we want to find the mass of lead (m) that would generate 3000 joules of energy. We know the specific heat capacity of aluminum (c), so we can use that as an approximation for lead. We also know the change in temperature (ΔT) from 100 °C to 40 °C.
Let's calculate:
- E = 3000 J
- c = 900 J/kg-°C
- ΔT = 100 °C - 40 °C = 60 °C
Now we can rearrange the formula to solve for the mass (m):
m = E / (c * ΔT)
Substituting the known values:
m = 3000 J / (900 J/kg-°C * 60 °C) = 0.0556 kg
Therefore, it would take approximately 0.0556 kilograms of lead to generate 3000 joules of energy when cooled from 100 °C to 40 °C.