Final answer:
The change in the thermal energy of a 10.0 g block of lead heated from 25°C to 30°C is calculated using the specific heat capacity of lead, resulting in an energy increase of 6.45 Joules.
Step-by-step explanation:
The question involves a calculation of the change in the thermal energy of a block of lead when its temperature is increased. Given the specific heat of lead, the mass of the lead block, and the temperature change, one can calculate the energy change using the formula:
ΔQ = m × c × ΔT
where:
- ΔQ is the change in thermal energy (in Joules),
- m is the mass of the object (in grams),
- c is the specific heat capacity (in J/g°C), and
- ΔT is the change in temperature (in °C).
For a 10.0 g block of lead with a specific heat of 0.129 J/g°C heated from 25°C to 30°C, the change in thermal energy ΔQ is calculated as follows:
ΔQ = (10.0 g) × (0.129 J/g°C) × (30°C - 25°C)
ΔQ = 10.0 g × 0.129 J/g°C × 5°C
ΔQ = 6.45 J
Therefore, the thermal energy of the lead block increases by 6.45 Joules.