To determine the mass of the lead sample, we will use the specific heat capacity formula:
[tex]\[ q = mc\Delta{T} \][/tex]
where:
- [tex]\( q \)[/tex] is the thermal energy in joules,
- [tex]\( m \)[/tex] is the mass in grams,
- [tex]\( c \)[/tex] is the specific heat capacity in J/g°C,
- [tex]\( \Delta{T} \)[/tex] is the change in temperature in degrees Celsius.
We were given that the specific heat of lead ([tex]\( c \)[/tex]) is 0.138 J/g°C, the energy released ([tex]\( q \)[/tex]) is 1.20 x 10^3 joules, the initial temperature is 93.0°C, and the final temperature is 29.5°C.
First, let's calculate the change in temperature ([tex]\( \Delta{T} \)[/tex]):
[tex]\[ \Delta{T} = \text{Initial Temperature} - \text{Final Temperature} \][/tex]
[tex]\[ \Delta{T} = 93.0°C - 29.5°C \][/tex]
[tex]\[ \Delta{T} = 63.5°C \][/tex]
Now that we have the change in temperature, we can rearrange the specific heat capacity formula to solve for the mass ([tex]\( m \)[/tex]):
[tex]\[ m = \frac{q}{c\Delta{T}} \][/tex]
Plug in the numbers:
[tex]\[ m = \frac{1.20 \times 10^3 \text{ J}}{0.138 \text{ J/g°C} \times 63.5°C} \][/tex]
Now perform the calculation:
[tex]\[ m = \frac{1.20 \times 10^3}{0.138 \times 63.5} \][/tex]
[tex]\[ m = \frac{1.20 \times 10^3}{8.769} \][/tex]
[tex]\[ m \approx 136.8 \text{ g} \][/tex]
So the mass of the lead sample is approximately 136.8 grams.
