If 10.0g of Pb at 25.0oC absorbs 1.00kJ of heat, what is it's final temperature?
38.3
To find the final temperature of Pb (Lead) after absorbing heat, we can use the equation:
q = mcΔT
where:
q is the heat absorbed (1.00 kJ)
m is the mass of the substance (10.0 g)
c is the specific heat capacity of the substance (which is known for lead)
ΔT is the change in temperature (final temperature - initial temperature)
First, let's convert the given values to appropriate units. The specific heat capacity (c) of lead is 0.129 J/g°C. We need to convert 1.00 kJ to joules by multiplying it by 1000, giving us 1000 J.
Now, we can rearrange the equation to solve for ΔT:
ΔT = q / (mc)
Substituting the given values:
ΔT = 1000 J / (10.0 g * 0.129 J/g°C)
Simplifying:
ΔT = 1000 J / 1.29 J/°C
ΔT = 775.19 °C
The change in temperature is 775.19 °C.
To find the final temperature, we add this change to the initial temperature of 25.0°C:
Final temperature = Initial temperature + ΔT
Final temperature = 25.0°C + 775.19 °C
Final temperature ≈ 800.19 °C
Therefore, the final temperature of Pb is approximately 800.19 °C after absorbing 1.00 kJ of heat.