A 15.0 gram lead ball at 25.0°C was heated with 40.5 joules of heat. Given the specific heat of lead is 0.128 J/g∙°C, what is the final temperature of the lead?
Answer = 46.10 celsius
Well, heating up a lead ball sounds like a lead-ing question! But don't worry, I'm here to give you an answer that's just as solid.
To find the final temperature of the lead ball, we can use the formula:
Q = mcΔT
Where Q is the heat energy, m is the mass of the lead ball, c is the specific heat of lead, and ΔT is the change in temperature.
In this case, we're trying to find the final temperature, so we rearrange the formula a bit:
ΔT = Q / (mc)
Now, let's plug in the values you've given:
mass (m) = 15.0 g
heat energy (Q) = 40.5 J
specific heat (c) = 0.128 J/g∙°C
ΔT = 40.5 J / (15.0 g * 0.128 J/g∙°C)
Now let me do a quick calculation to find ΔT...
*beep boop beep*
Ah, I've got it! The change in temperature (ΔT) is approximately 21.09 °C.
To find the final temperature, we'll need to add this change to the initial temperature. Since the initial temperature is given as 25.0 °C, we just add ΔT to it:
Final temperature = Initial temperature + ΔT
Final temperature = 25.0 °C + 21.09 °C
Doing some more math...
*beep boop beep*
The final temperature of the lead ball is approximately 46.09 °C. Voilà!
Hope that gives you a clear lead on the answer!
To find the final temperature of the lead ball, we can use the formula for heat transfer:
Q = mcΔT
Where:
Q is the heat transferred (40.5 J)
m is the mass of the lead ball (15.0 g)
c is the specific heat of lead (0.128 J/g∙°C)
ΔT is the change in temperature
First, let's rearrange the formula to solve for ΔT:
ΔT = Q / (mc)
ΔT = 40.5 J / (15.0 g * 0.128 J/g∙°C)
Next, we calculate the value of ΔT:
ΔT = 40.5 J / (1.92 g∙°C)
Simplifying the calculation:
ΔT ≈ 21.094 °C
Finally, we find the final temperature by adding the change in temperature to the initial temperature:
Final temperature = Initial temperature + ΔT = 25.0 °C + 21.094 °C
Final temperature ≈ 46.094 °C
Therefore, the final temperature of the lead ball is approximately 46.094 °C.
To find the final temperature of the lead, we can use the formula for heat transfer:
q = m * c * ΔT
where:
q is the heat transferred (in joules)
m is the mass of the substance (in grams)
c is the specific heat capacity of the substance (in J/g∙°C)
ΔT is the change in temperature of the substance (in °C)
In this case, we know:
q = 40.5 J
m = 15.0 g
c = 0.128 J/g∙°C
Rearranging the formula, we have:
ΔT = q / (m * c)
Plugging in the given values:
ΔT = 40.5 J / (15.0 g * 0.128 J/g∙°C)
Now we can calculate the change in temperature:
ΔT ≈ 26.56 °C
The change in temperature represents the increase in temperature from the initial temperature to the final temperature. To find the final temperature, we need to add the change in temperature to the initial temperature:
Final temperature = Initial temperature + ΔT
Given that the initial temperature is 25.0 °C:
Final temperature = 25.0 °C + 26.56 °C
Final temperature ≈ 51.56 °C
Therefore, the final temperature of the lead is approximately 51.56 °C.