hi this seems simple but i cant figure it out.
If a 0.46 kg hot copper pan lost 9.8x103
J of heat as it cooled down to 54°C, what was its
original temperature?
I used the 24.44 j/molK for heat capacity, is not this one? found on wikipedia: copper
this is how i derive my equation...
-9.8x10^3=7.24mol(24.44j/molK)(327-x)
the answer is 77C but im getting 109.39C :( please help thanks
I used 0.385 J/g*C and came up with the same answer. You can prove that 109.39 is correct (although that's too many significant figures to use) this way.
109-54 = 55.39
55.39 x 0.385 J/g x 460 g = 980.9 J.
55.39 x 24.44 x (460/63.54) = 980.03 J.
thanks!!!! it was an exercise from study guide it just gave the answer not the solution. might be correction
To solve this problem, we can first use the equation:
Q = mcΔT
where Q is the heat lost, m is the mass of the pan, c is the specific heat capacity of copper, and ΔT is the change in temperature.
In this case, the mass of the pan (m) is given as 0.46 kg, the specific heat capacity of copper (c) is given as 24.44 J/(mol·K), and the change in temperature (ΔT) is the difference between the final temperature (54°C) and the original temperature (let's call it T).
So we have the equation:
-9.8x10^3 J = (0.46 kg)(24.44 J/(mol·K))(54°C - T)
Now, let's solve for T.
First, convert the temperature to Kelvin by adding 273.15 to each side:
54°C + 273.15 = T + 273.15
Now the equation becomes:
-9.8x10^3 J = (0.46 kg)(24.44 J/(mol·K))(327.15 K - T)
Simplifying the equation:
-9.8x10^3 J = 7.24 mol(24.44 J/(mol·K))(327.15 K - T)
Now divide both sides of the equation by (7.24 mol)(24.44 J/(mol·K)):
-9.8x10^3 J / (7.24 mol)(24.44 J/(mol·K)) = 327.15 K - T
Simplifying:
-9.8x10^3 J / (7.24 mol)(24.44 J/(mol·K)) ≈ 77 K - T
Finally, subtract 77 K from both sides:
-9.8x10^3 J / (7.24 mol)(24.44 J/(mol·K)) - 77 K ≈ -T
T ≈ 77 K (approx.)
Therefore, the original temperature of the copper pan is approximately 77°C.