Burrito Bunny wrote:I HAVE A QUESTION!
TYPE OF QUESTION: Chemistry
YOUR QUESTION:
We did an experiment where first we heated water in a beamer, and then placed a rock into that beaker for 3 minutes. After taking the temperature of the rock while I'm the hot water, we removed the rock and placed it into a calorimeter, tracking the heat until we found the temperature the heated rock and room temperature water within reached equilibrium. Here's some data:Mass of water in the calorimeter: 97.87g
Mass of the rock: 113.74g
Initial temperature of the rock (in the hot water): 84.6°C
Initial temperature of the water in calorimeter: 22.0°C
Final temperature of rock and water in calorimeter: 34.2°C
ΔT of water in calorimeter: 12.2°C
ΔT of rock: 50.4 °C
Specific Heat (Cp) of rock: 0.86 J/g °C
Soooo, I'm having trouble with this final problem. It asks me to show math but I'm not sure how to get to the answer. Here's the question:The specific heat (Cp) of lead is 0.13 joules/g°C.
If you had used 50g of lead instead of the rock in this experiment, would the temperature of the water be more or less than you found? Explain why! (Show your calculations).
I'm pretty sure it wants me to use the equation Q = m • Cp • ΔT. But...i seem to be missing Q and ΔT so I'm not sure how to find the missing variable! I thought maybe I was supposed to assume the initial temperature of the rock and the lead were the same, but even then I'm not sure of the final temperature so I'm unsure of ΔT' s value.
Any help would be greatly appriciated!!!
I'm a little confused at the bolded because you straight up have a ΔT recorded in your data? I think you might be overthinking this bit a little. o.o And if you're just missing Q, then you can solve for Q.
Just looking over the question, let's use a little logic first. The specific heat of lead is much less than the specific heat of rock (way less than half), while the proposed amount of lead used is half of what you used for the rock. That's two of your multiplication variables that have just decreased, so it stands to follow that if you're solving for Q, then the temp of the water of should be less than what you found in your experiment. Right?
Even if you can't complete a calculation, write down your thoughts and reasons - a lot of teachers will give you partial credit for knowing a solid portion of what you were supposed to do, even if you couldn't complete it!
Someone else who comes along might have done chemistry more recently than me (eugh, it's been a year now), though, and have a more helpful answer.