A piece of lead with a mass of 27.3 g was heated to 98.90oC and then dropped into 15,0 g of water at 22.50oC. The final temperature was 26.32oC. Calculate the specific heat capacity of lead from these data.
This is what I did. Could you check my answer? I would really appreciate it.
-qlead=qwater
q = mcΔt
-mcΔt = mcΔt
-(27.3 g)(clead)(26.32-98.90)=(15.0g)(4.184)(26.32-22.50)
clead = .121
I obtained 0.12088 which rounds to 0.121 (your answer) if we have only three significant figures; that is, if the mass of Pb is 27.3 g and not 27.30 and the mass of the water is 15.0 g and not 15.00
Your method is correct.
I did not check the arithmetic but my physics book says lead is .130 so I think you did it right.
Thank you both very much for your help and for responding as quickly as you did!
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To calculate the specific heat capacity of lead, you used the equation qlead = qwater, which is correct.
Now, let's break down your calculations:
1. First, you used the equation q = mcΔt, where q represents the heat absorbed or released, m is the mass, c is the specific heat capacity, and Δt is the change in temperature.
2. Then, you equated the heat absorbed by the lead with the heat released by the water: -(27.3 g)(clead)(26.32-98.90)=(15.0g)(4.184)(26.32-22.50).
3. Solving for clead, you obtained clead = 0.121.
It seems like you made a mistake in your calculation. Let's correct it:
-(27.3 g)(clead)(26.32-98.90)=(15.0g)(4.184)(26.32-22.50)
(-27.3 g)(clead)(-72.58) = (15.0 g)(4.184)(3.82)
clead = [(15.0 g)(4.184)(3.82)] / [(27.3 g)(72.58)]
clead = 0.1217
Therefore, the specific heat capacity of lead is approximately 0.1217 J/g°C.
Your answer was close; you just made a calculation error.