What is the specific heat of lead if 13.6 cal are needed to raise the temperature of 35.6 g Pb by 12.5 LaTeX: ^\circ∘ C ? Your answer should have 3 sig figs.
_____cal / g °C.
You missed the whole point. All you need to do is to substitute into the equation I gave you and you didn't do that.
q = mass x specific heat x delta T
q = 13.6 calories from the problem.
mass Pb = 35.6 g from the problem.
specific heat of the Pb is what the problem is asking you to calculate. That's the unknown in the equation.
delta T is 12.5 C from the problem. So
13.6 cal = 35.6 g x specific heat x 12.5 C
Solve for specific heat. I looked up the specific heat Pb on the web and found 0.0301 cal/g*C. You may or may not get that answer but that's the way to work the problem. Make sure I substituted the correct values from the problem into the equation.
13.6 cal = mass Pb x specific heat Pb x delta T
DrBob222
so it will be like 13.6 = 35.6(12.5)(13.6)?
I feel like I am missing something.
To find the specific heat (cal/g °C) of lead, we can use the formula:
specific heat = (heat absorbed / mass * change in temperature)
Given:
Heat absorbed = 13.6 cal
Mass = 35.6 g
Change in temperature = 12.5 °C
Using the formula, we can substitute the given values to find the specific heat:
specific heat = (13.6 cal) / (35.6 g * 12.5 °C)
Calculating this expression, we get:
specific heat = 0.03104225 cal/g °C
Rounding this value to 3 significant figures, the specific heat of lead is approximately 0.031 cal/g °C.
Therefore, the answer is:
0.031 cal/g °C