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
mPb = 50g (0.05 kg)
Q = 8.7 * 10^2 J
T = (x-22)
x being the final temperature
870 = (0.05 kg)(1.3*10^2)(x-22)
870 = 6.5(x-22)
870 = 6.5x - 143
870 + 143 = 6.5x
1013 = 6.5x
Divide both sides by 6.5
x = 155.8 (Round sig digs to 2 places)
x = 160
Well, let's just say that lead sure knows how to warm up! To find the final temperature, we can make use of the specific heat formula. The specific heat of lead is 0.128 J/g°C.
First, let's calculate the heat gained using the formula Q = mcΔT, where Q is the energy absorbed, m is the mass, c is the specific heat, and ΔT is the change in temperature.
The mass of the lead is 50.0 g, the specific heat is 0.128 J/g°C, and the energy absorbed is 8.7 X 10^2 J.
So we have:
8.7 X 10^2 J = (50.0 g) * (0.128 J/g°C) * ΔT
Now, we can solve for ΔT by dividing both sides of the equation by (50.0 g) * (0.128 J/g°C):
ΔT = (8.7 X 10^2 J) / (50.0 g * 0.128 J/g°C)
Calculating that out, we find:
ΔT ≈ 135.94 °C
Finally, to find the final temperature, we add the change in temperature to the initial temperature:
Final temperature = 22 °C + 135.94 °C
You can grab your calculator and sum it up to find the final temperature!
To find the final temperature of the lead, we can use the formula:
Q = mcΔT
Where:
Q = amount of energy absorbed by the lead (in joules)
m = mass of the lead (in grams)
c = specific heat capacity of lead (in joules/gram·degree Celsius)
ΔT = change in temperature (in degrees Celsius)
First, let's convert the mass of lead from grams to kilograms:
m = 50.0 g = 0.050 kg
Next, let's find the specific heat capacity of lead. The specific heat capacity of lead is approximately 0.13 joules/gram·degree Celsius.
Now, rearranging the formula, we can solve for ΔT:
ΔT = Q / (mc)
Substituting the given values:
Q = 8.7 × 10^2 J
m = 0.050 kg
c = 0.13 J/g·°C
ΔT = (8.7 × 10^2 J) / (0.050 kg × 0.13 J/g·°C)
ΔT = (8.7 × 10^2 J) / (0.0065 J/°C)
Evaluating the expression:
ΔT = 133.846 °C
Finally, we need to add the change in temperature to the initial temperature to find the final temperature:
Final temperature = Initial temperature + ΔT
Final temperature = 22 °C + 133.846 °C
Final temperature = 155.846 °C
Therefore, the final temperature of the lead is approximately 155.846 °C.
To find the final temperature of the lead, we need to use the specific heat capacity formula:
q = mcΔT
where:
q is the amount of energy absorbed,
m is the mass of the substance (lead in this case),
c is the specific heat capacity of the substance (in this case, we'll assume it is the specific heat capacity of lead), and
ΔT is the change in temperature.
First, let's convert the energy absorbed from scientific notation to standard form:
8.7 x 10^2 = 870
The specific heat capacity of lead is approximately 0.13 J/g°C.
Now, we rearrange the formula to solve for ΔT:
ΔT = q / (mc)
Substituting the known values:
ΔT = 870 J / (50.0 g * 0.13 J/g°C)
ΔT = 870 J / 6.50 g°C
ΔT ≈ 134.15 °C
Now, to find the final temperature, we add the change in temperature to the initial temperature:
Final temperature = Initial temperature + ΔT
Final temperature = 22°C + 134.15 °C
Final temperature ≈ 156.15 °C
Therefore, the final temperature of the lead is approximately 156.15 °C.