A 50.0 g sample of lead starts at 22 degrees Celsius and is heated until it absorbs 8.7 X 10^2 of energy. Find the final temperature of the lead
To find the final temperature of the lead sample, we can use the specific heat formula:
q = mcΔT
Where:
q = the amount of energy absorbed by the lead (in joules)
m = mass of the lead (in grams)
c = specific heat of lead (in J/g°C)
ΔT = change in temperature (in °C)
First, let's convert the mass of the lead from grams to kilograms:
50.0 g = 0.0500 kg
Next, let's calculate the change in temperature by rearranging the formula:
ΔT = q / (mc)
Given:
q = 8.7 x 10^2 J
m = 0.0500 kg
c = specific heat of lead (since it's not provided, we can use the approximate value of 0.13 J/g°C for lead)
ΔT = (8.7 x 10^2 J) / (0.0500 kg x 0.13 J/g°C)
Now, let's perform the calculations:
ΔT = (8.7 x 10^2) / (0.0500 x 0.13)
= (870) / (0.0065)
= 133846.15 °C
Finally, to find the final temperature, we add the change in temperature to the initial temperature:
Final temperature = Initial temperature + ΔT
Final temperature = 22°C + 133846.15°C
Final temperature ≈ 133868.15°C
Therefore, the final temperature of the lead sample after absorbing 8.7 x 10^2 J of energy is approximately 133868.15°C.
first, pretend it does not get to the melting point.
8700J=50*specificheat*(tf-22)
8700=50*.160*(Tf-22)
Tf-22= 1087 so it melted...
OK, figure how much eat it took to raised it to melting point.
Q=50*.160*(621-22)
I get about 4300 J (you check that)
so you have left44ooJ
Now how hot does it get
4400=50*(sjpecific heat moltenlead(Tf-621)
Tf-621=4400/(50*.150)=587
so final temp is in the range of 1200C
You do all that accurately.